================ Start Lecture #1 ================
G22.2130: Compiler Construction
2006-07 Spring
Allan Gottlieb
Tuesday 5-6:50pm Rm 109 Ciww

Chapter 0: Administrivia

I start at Chapter 0 so that when we get to chapter 1, the numbering will agree with the text.

0.1: Contact Information

0.2: Course Web Page

There is a web site for the course. You can find it from my home page listed above.

0.3: Textbook

The course text is Aho, Lam, Seithi, and Ullman: Compilers: Principles, Techniques, and Tools, second edition

0.4: Computer Accounts and Mailman Mailing List

0.5: Grades

Your grade will be a function of your final exam and laboratory assignments (see below). I am not yet sure of the exact weightings for each lab and the final, but the final will be roughly half the grade (very likely between 40% and 60%).

0.6: The Upper Left Board

I use the upper left board for lab/homework assignments and announcements. I should never erase that board. If you see me start to erase an announcement, please let me know.

I try very hard to remember to write all announcements on the upper left board and I am normally successful. If, during class, you see that I have forgotten to record something, please let me know. HOWEVER, if I forgot and no one reminds me, the assignment has still been given.

0.7: Homeworks and Labs

I make a distinction between homeworks and labs.

Labs are

Homeworks are

0.7.1: Homework Numbering

Homeworks are numbered by the class in which they are assigned. So any homework given today is homework #1. Even if I do not give homework today, the homework assigned next class will be homework #2. Unless I explicitly state otherwise, all homeworks assignments can be found in the class notes. So the homework present in the notes for lecture #n is homework #n (even if I inadvertently forgot to write it to the upper left board).

0.7.2: Doing Labs on non-NYU Systems

You may solve lab assignments on any system you wish, but ...

0.7.3: Obtaining Help with the Labs

Good methods for obtaining help include

  1. Asking me during office hours (see web page for my hours).
  2. Asking the mailing list.
  3. Asking another student, but ...
    Your lab must be your own.
    That is, each student must submit a unique lab. Naturally, simply changing comments, variable names, etc. does not produce a unique lab.

0.7.4: Computer Language Used for Labs

You may write your lab in Java, C, or C++. Other languages may be possible, but please ask in advance. I need to ensure that the TA is comfortable with the language.

0.8: A Grade of “Incomplete”

The rules for incompletes and grade changes are set by the school and not the department or individual faculty member. The rules set by GSAS state:

The assignment of the grade Incomplete Pass(IP) or Incomplete Fail(IF) is at the discretion of the instructor. If an incomplete grade is not changed to a permanent grade by the instructor within one year of the beginning of the course, Incomplete Pass(IP) lapses to No Credit(N), and Incomplete Fail(IF) lapses to Failure(F).

Permanent grades may not be changed unless the original grade resulted from a clerical error.

0.9: An Introductory Compiler Course with a Programming Prerequisite

0.9.1: This is an introductory course ...

I do not assume you have had a compiler course as an undergraduate, and I do not assume you have had experience developing/maintaining a compiler.

If you have already had a compiler class, this course is probably not appropriate. For example, if you can explain the following concepts/terms, the course is probably too elementary for you.

... with a Programming Prerequisite

I do assume you are an experienced programmer. There will be non-trivial programming assignments during this course. Indeed, you will write a compiler for a simple programming language.

I also assume that you have at least a passing familiarity with assembler language. In particular, your compiler may need to produce assembler language. We will also be using addressing modes found in typical assemblers. We will not, however, write significant assembly-language programs.

0.10: Academic Integrity Policy

Our policy on academic integrity, which applies to all graduate courses in the department, can be found here.

Roadmap of the Course

  1. Chapter 1 touches on all the material.

  2. Chapter 2 constructs (the front end of) a simple compiler.

  3. Chapters 3-8 fill in the (considerable) gaps, as well as the beginnings of the back end.

  4. I tend to spend too much time on introductory chapters, but will try not to.

Chapter 1: Introduction to Compiling

Homework Read chapter 1.

1.1: Language Processors

A Compiler is a translator from one language, the input or source language, to another language, the output or target language.

Often, but not always, the target language is an assembler language or the machine language for a computer processor.

Note that using a compiler requires a two step process to run a program.

  1. Execute the compiler (and possibly an assembler) to translate the source program into a machine language program.
  2. Execute the resulting machine language program, supplying appropriate input.

This should be compared with an interpreter, which accepts the source language program and the appropriate input, and itself produces the program output.

Sometimes both compilation and interpretation are used. For example, consider typical Java implementations. The (Java) source code is translated (i.e., compiled) into bytecodes, the machine language for an idealized virtual machine, the Java Virtual Machine or JVM. Then an interpreter of the JVM (itself normally called a JVM) accepts the bytecodes and the appropriate input, and produces the output. This technique was quite popular in academia, with the Pascal programming language and P-code.

The compilation tool chain

For large programs, the compiler is actually part of a multistep tool chain

[preprocessor] → [compiler] → [assembler] → [linker] → [loader]

We will be primarily focused on the second element of the chain, the compiler. Our target language will be assembly language. I give a very short description of the other components, including some historical comments.

Preprocessors

Preprocessors are normally fairly simple as in the C language, providing primarily the ability to include files and expand macros. There are exceptions, however. IBM's PL/I, another Algol-like language had quite an extensive preprocessor, which made available at preprocessor time, much of the PL/I language itself (e.g., loops and I believe procedure calls).

Some preprocessors essentially augment the base language, to add additional capabilities. One could consider them as compilers in their own right, having as source this augmented language (say Fortran augmented with statements for multiprocessor execution in the guise of Fortran comments) and as target the original base language (in this case Fortran). Often the “preprocessor” inserts procedure calls to implement the extensions at runtime.

Assemblers

Assembly code is an mnemonic version of machine code in which names, rather than binary values, are used for machine instructions, and memory addresses.

Some processors have fairly regular operations and as a result assembly code for them can be fairly natural and not-too-hard to understand. Other processors, in particular Intel's x86 line, have let us charitably say more interesting instructions with certain registers used for certain things.

My laptop has one of these latter processors (pentium 4) so my gcc compiler produces code that from a pedagogical viewpoint is less than ideal. If you have a mac with a ppc processor (newest macs are x86), your assembly language is cleaner. NYU's ACF features sun computers with sparc processors, which also have regular instruction sets.

Two pass assembly

No matter what the assembly language is, an assembler needs to assign memory locations to symbols (called identifiers) and use the numeric location address in the target machine language produced. Of course the same address must be used for all occurrences of a given identifier and two different identifiers must (normally) be assigned two different locations.

The conceptually simplest way to accomplish this is to make two passes over the input (read it once, then read it again from the beginning). During the first pass, each time a new identifier is encountered, an address is assigned and the pair (identifier, address) is stored in a symbol table. During the second pass, whenever an identifier is encountered, its address is looked up in the symbol table and this value is used in the generated machine instruction.

Linkers

Linkers, a.k.a. linkage editors combine the output of the assembler for several different compilations. That is the horizontal line of the diagram above should really be a collection of lines converging on the linker. The linker has another input, namely libraries, but to the linker the libraries look like other programs compiled and assembled. The two primary tasks of the linker are

  1. Relocating relative addresses.
  2. Resolving external references (such as the procedure xor() above).
Relocating relative addresses

The assembler processes one file at a time. Thus the symbol table produced while processing file A is independent of the symbols defined in file B, and conversely. Thus, it is likely that the same address will be used for different symbols in each program. The technical term is that the (local) addresses in the symbol table for file A are relative to file A; they must be relocated by the linker. This is accomplished by adding the starting address of file A (which in turn is the sum of the lengths of all the files processed previously in this run) to the relative address.

Resolving external references

Assume procedure f, in file A, and procedure g, in file B, are compiled (and assembled) separately. Assume also that f invokes g. Since the compiler and assembler do not see g when processing f, it appears impossible for procedure f to know where in memory to find g.

The solution is for the compiler to indicated in the output of the file A compilation that the address of g is needed. This is called a use of g. When processing file B, the compiler outputs the (relative) address of g. This is called the definition of g. The assembler passes this information to the linker.

The simplest linker technique is to again make two passes. During the first pass, the linker records in its “external symbol table” (a table of external symbols, not a symbol table that is stored externally) all the definitions encountered. During the second pass, every use can be resolved by access to the table.

I will be covering the linker in more detail tomorrow at 5pm in 2250, OS Design

Loaders

After the linker has done its work, the resulting “executable file” can be loaded by the operating system into central memory. The details are OS dependent. With early single-user operating systems all programs would be loaded into a fixed address (say 0) and the loader simply copies the file to memory. Today it is much more complicated since (parts of) many programs reside in memory at the same time. Hence the compiler/assembler/linker cannot know the real location for an identifier. Indeed, this real location can change.

More information is given in any OS course (e.g., 2250 given wednesdays at 5pm).

Homework: 1, 4

Remark
Unless state otherwise, homeworks are from the book and specifically from the end of the second level section we are discussing. Even more specifically, we are in section 1.1, so you are to do the first and fourth problem at the end of section 1.1. These two problems are numbered 1.1.1 and 1.1.4 in the book.
End of Remark phases

1.2: The Structure of a Compiler

Modern compilers contain two (large) parts, each of which is often subdivided. These two parts are the front end, shown in green on the right and the back end, shown in pink.

The front end analyzes the source program, determines its constituent parts, and constructs an intermediate representation of the program. Typically the front end is independent of the target language.

The back end synthesizes the target program from the intermediate representation produced by the front end. Typically the back end is independent of the source language.

This front/back division very much reduces the work for a compiling system that can handle several (N) source languages and several (M) target languages. Instead of NM compilers, we need N front ends and M back ends. For gcc (originally standing for Gnu C Compiler, but now standing for Gnu Compiler Collection), N=7 and M~30 so the savings is considerable.

Multiple Phases

The front and back end are themselves each divided into multiple phases. The input to each phase is the output of the previous. Sometime a phase changes the representation of the input. For example, the lexical analyzer converts a character stream input into a token stream output. Sometimes the representation is unchanged. For example, the machine-dependent optimizer transforms target-machine code into (hopefully improved) target-machine code.

The diagram is definitely not drawn to scale, in terms of effort or lines of code. In practice the optimizers, especially the machine-dependent one, dominate.

Conceptually, there are three phases of analysis with the output of one phase the input of the next. The phases are called lexical analysis or scanning, syntax analysis or parsing, and semantic analysis.

1.2.1: Lexical Analysis (or Scanning)

The character stream input is grouped into meaningful units called lexemes, which are then mapped into tokens, the latter constituting the output of the lexical analyzer. For example, any one of the following

  x3 = y + 3;
  x3  =   y   +   3   ;
  x3   =y+ 3  ;
but not
  x 3 = y + 3;
would be grouped into the lexemes x3, =, y, +, 3, and ;.

A token is a <token-name,attribute-value> pair. For example

  1. The lexeme x3 would be mapped to a token such as <id,1>. The name id is short for identifier. The value 1 is the index of the entry for x3 in the symbol table produced by the compiler. This table is used to pass information to subsequent phases.
  2. The lexeme = would be mapped to the token <=>. In reality it is probably mapped to a pair, whose second component is ignored. The point is that there are many different identifiers so we need the second component, but there is only one assignment symbol =.
  3. The lexeme y is mapped to the token <id,2>
  4. The lexeme + is mapped to the token <+>.
  5. The lexeme 3 is somewhat interesting and is discussed further in subsequent chapters. It is mapped to <number,something>, but what is the something. On the one hand there is only one 3 so we could just use the token <number,3>. However, there can be a difference between how this should be printed (e.g., in an error message produced by subsequent phases) and how it should be stored (fixed vs. float vs double). Perhaps the token should point to the symbol table where an entry for this kind of 3 is stored. Another possibility is to have a separate numbers table.
  6. The lexeme ; is mapped to the token <;>.

Note that non-significant blanks are normally removed during scanning. In C, most blanks are non-significant. Blanks inside strings are an exception.

Note that we can define identifiers, numbers, and the various symbols and punctuation without using recursion (compare with parsing below). parse-tree

1.2.2: Syntax Analysis (or Parsing)

Parsing involves a further grouping in which tokens are grouped into grammatical phrases, which are often represented in a parse tree. For example

    x3 = y + 3;
  
would be parsed into the tree on the right.

This parsing would result from a grammar containing rules such as

    asst-stmt → id = expr ;
    expr      → number
              |  id
              |  expr + expr
  

Note the recursive definition of expression (expr). Note also the hierarchical decomposition in the figure on the right.

The division between scanning and parsing is somewhat arbitrary, but invariably if a recursive definition is involved, it is considered parsing not scanning. syntax tree

Often we utilize a simpler tree called the syntax tree with operators as interior nodes and operands as the children of the operator. The syntax tree on the right corresponds to the parse tree above it.

(Technical point.) The syntax tree represents an assignment expression not an assignment statement. In C an assignment statement includes the trailing semicolon. That is, in C (unlike in Algol) the semicolon is a statement terminator not a statement separator.

1.2.3: Semantic Analysis

semantic-tree

There is more to a front end than simply syntax. The compiler needs semantic information, e.g., the types (integer, real, pointer to array of integers, etc) of the objects involved. This enables checking for semantic errors and inserting type conversion where necessary.

For example, if y was declared to be a real and x3 an integer, we need to insert (unary, i.e., one operand) conversion operators “inttoreal” and “realtoint” as shown on the right.

1.2.4: Intermediate code generation

Many compilers first generate code for an “idealized machine”. For example, the intermediate code generated would assume that the target has an unlimited number of registers and that any register can be used for any operation. Another common assumption is that machine operations take (up to) three operands, two source and one target.

With these assumptions one generates three-address code by walking the semantic tree. Our example C instruction would produce

temp1 = inttoreal(3)
temp2 = id2 + temp1
temp3 = realtoint(temp2)
id1 = temp3

We see that three-address code can include instructions with fewer than 3 operands.

Sometimes three-address code is called quadruples because one can view the previous code sequence as

inttoreal temp1 3     --
add       temp2 id2   temp1
realtoint temp3 temp2 --
assign    id1   temp3 --
Each “quad” has the form
  operation  target source1 source2

1.2.5: Code optimization

This is a very serious subject, one that we will not really do justice to in this introductory course. Some optimizations are fairly easy to see.

  1. Since 3 is a constant, the compiler can perform the int to real conversion and replace the first two quads with
      add       temp2 id2  3.0
      
  2. The last two quads can be combined into
      realtoint id1   temp2
      

1.2.6: Code generation

Modern processors have only a limited number of register. Although some processors, such as the x86, can perform operations directly on memory locations, we will for now assume only register operations. Some processors (e.g., the MIPS architecture) use three-address instructions. However, some processors permit only two addresses; the result overwrites one of the sources. With these assumptions, code something like the following would be produced for our example, after first assigning memory locations to id1 and id2.

    LD   R1,  id2
    ADDF R1,  R1, #3.0    // add float
    RTOI R2,  R1          // real to int
    ST   id1, R2
  

1.2.7: Symbol-Table Management

The symbol table stores information about program variables that will be used across phases. Typically, this includes type information and storage location.

A possible point of confusion: the storage location does not give the location where the compiler has stored the variable. Instead, it gives the location where the compiled program will store the variable.

1.2.8: The Grouping of Phases into Passes

Logically each phase is viewed as a separate program that reads input and produces output for the next phase, i.e., a pipeline. In practice some phases are combined into a pass.

For example one could have the entire front end as one pass.

The term pass is used to indicate that the entire input is read during this activity. So two passes, means that the input is read twice. We have discussed two pass approaches for both assemblers and linkers. If we implement each phase separately and use multiple passes for some of them, the compiler will perform a large number of I/O operations, an expensive undertaking.

As a result techniques have been developed to reduce the number of passes. We will see in the next chapter how to combine the scanner, parser, and semantic analyzer into one phase. Consider the parser. When it needs another token, rather than reading the input file (presumably produced by the scanner), the parser calls the scanner instead. At selected points during the production of the syntax tree, the parser calls the intermediate-code generator which performs semantic analysis as well as generating a portion of the intermediate code.

For pedagogical reasons, we will not be employing this technique. Thus your compiler will consist of separate programs for the scanner, parser, and semantic analyzer / intermediate code generator. Indeed, these will very likely be labs 2, 3, and 4.

Reducing the number of passes

One problem with combining phases, or with implementing a single phase in one pass, is that it appears that an internal form of the entire program will need to be stored in memory. This problem arises because the downstream phase may need early in its execution, information that the upstream phase produces only late in its execution. This motivates the use of symbol tables and a two pass approach. However, a clever one-pass approach is often possible.

Consider the assembler (or linker). The good case is when the definition precedes all uses so that the symbol table contains the value of the symbol prior to that value being needed. Now consider the harder case of one or more uses preceding the definition. When a not-yet-defined symbol is first used, an entry is placed in the symbol table, pointing to this use and indicating that the definition has not yet appeared. Further uses of the same symbol attach their addresses to a linked list of “undefined uses” of this symbol. When the definition is finally seen, the value is placed in the symbol table, and the linked list is traversed inserting the value in all previously encountered uses. Subsequent uses of the symbol will find its definition in the table.

This technique is called backpatching.

1.2.9: Compiler-construction tools

Originally, compilers were written “from scratch”, but now the situation is quite different. A number of tools are available to ease the burden.

We will study tools that generate scanners and parsers. This will involve us in some theory, regular expressions for scanners and various grammars for parsers. These techniques are fairly successful. One drawback can be that they do not execute as fast as “hand-crafted” scanners and parsers.

We will also see tools for syntax-directed translation and automatic code generation. The automation in these cases is not as complete.

Finally, there is the large area of optimization. This is not automated; however, a basic component of optimization is “data-flow analysis” (how values are transmitted between parts of a program) and there are tools to help with this task.

1.3: The Evolution of Programming Languages

1.3.1: The Move to Higher-level Languages

Skipped. Assumed knowledge (only one page).

1.3.2: Impacts on Compilers

High performance compilers (i.e., the code generated performs well) are crucial for the adoption of new language concepts and computer architectures. Also important is the resource utilization of the compiler itself.

Modern compilers are large. On my laptop the compressed source of gcc is 38MB so uncompressed it must be about 100MB.

1.4: The Science of Building a Compiler

1.4.1: Modeling in Compiler Design and Implementation

We will encounter several aspects of computer science during the course. Some, e.g., trees, I'm sure you already know well. Other, more theoretical aspects, such as nondeterministic finite automata, may be new.

1.4.2: The Science of Code Optimization

We will do very little optimization. That topic is typically the subject of a second compiler course. Considerable theory has been developed for optimization, but sadly we will see essentially none of it. We can, however, appreciate the pragmatic requirements.

1.5: Applications of Compiler Technology

1.5.1: Implementation of High-Level Programming Languages

1.5.2: Optimization for Computer Architectures

Parallelism

Major research efforts had lead to improvements in

Memory Hierarchies

All machines have a limited number of registers, which can be accessed much faster than central memory. All but the simplest compilers devote effort to using this scarce resource effectively. Modern processors have several levels of caches and advanced compilers produce code designed to utilize the caches well.

1.5.3: Design of New Computer Architectures

RISC (Reduced Instruction Set Computer)

RISC computers have comparatively simple instructions, complicated instructions require several RISC instructions. A CISC, Complex Instruction Set Computer, contains both complex and simple instructions. A sequence of CISC instructions would be a larger sequence of RISC instructions. Advanced optimizations are able to find commonality in this larger sequence and lower the total number of instructions. The CISC Intel x86 processor line 8086/80286/80386/... had a major change with the 686 (a.k.a. pentium pro). In this processor, the CISC instructions were decomposed into RISC instructions by the processor itself. Currently, code for x86 processors normally achieves highest performance when the (optimizing) compiler emits primarily simple instructions.

Specialized Architectures

A great variety has emerged. Compilers are produced before the processors are fabricated. Indeed, compilation plus simulated execution of the generated machine code is used to evaluate proposed designs.

1.5.4: Program Translations

Binary Translation

This means translating from one machine language to another. Companies changing processors sometimes use binary translation to execute legacy code on new machines. Apple did this when converting from Motorola CISC processors to the PowerPC. An alternative is to have the new processor execute programs in both the new and old instruction set. Intel had the Itanium processor also execute x86 code. Apple, however, did not produce their own processors.

With the recent dominance of x86 processors, binary translators from x86 have been developed so that other microprocessors can be used to execute x86 software.

Hardware Synthesis

In the old days integrated circuits were designed by hand. For example, the NYU Ultracomputer research group in the 1980s designed a VLSI chip for rapid interprocessor coordination. The design software we used essentially let you paint. You painted blue lines where you wanted metal, green for polysilicon, etc. Where certain colors crossed, a transistor appeared.

Current microprocessors are much too complicated to permit such a low-level approach. Instead, designers write in a high level description language which is compiled down the specific layout.

Database Query Interpreters

The optimization of database queries and transactions is quite a serious subject.

Compiled Simulation

Instead of simulating a designs on many inputs, it may be faster to compiler the design first into a lower level representation and then execute the compiled version.

1.5.5: Software Productivity Tools

Dataflow techniques developed for optimizing code are also useful for finding errors. Here correctness is not an absolute requirement, a good thing since finding all errors in undecidable.

Type Checking

Techniques developed to check for type correctness (we will see some of these) can be extended to find other errors such as using an uninitialized variable.

Bounds Checking

As mentioned above optimizations have been developed to eliminate unnecessary bounds checking for languages like Ada and Java that perform the checks automatically. Similar techniques can help find potential buffer overflow errors that can be a serious security threat.

Memory-Management Tools

Languages (e.g., Java) with garbage collection cannot have memory leaks (failure to free no longer accessible memory). Compilation techniques can help to find these leaks in languages like C that do not have garbage collection.

1.6: Programming Language Basics

Skipped (prerequisite).

================ Start Lecture #2 ================

Chapter 2: A Simple Syntax-Directed Translator

Remark: I had a bad link to the mailing list. It was fixed on wed about 1:45pm. If you signed up before then you might have joined last year's list. Sorry.

Homework: Read chapter 2.

The goal of this chapter is to implement a very simple compiler. Really we are just going as far as the intermediate code, i.e., the front end. Nonetheless, the output, i.e. the intermediate code, does look somewhat like assembly language

  1. Simple source language.
  2. No optimization.
  3. No machine-dependent back end.
  4. No tools.
  5. Little theory.

2.1: Introduction

We will be looking at the front end, i.e., the analysis portion of a compiler.

The syntax describes the form of a program in a given language, while the semantics describes the meaning of that program. We will use the standard context-free grammar or BNF (Backus-Naur Form) to describe the syntax

We will learn syntax-directed translation, where the grammar does more than specify the syntax. We augment the grammar with attributes and use this to guide the entire front end.

The front end discussed in this chapter has as source language infix expressions consisting of digits, +, and -. The target language is postfix expressions with the same components. The compiler will convert
7+4-5 to 74+5-.
Actually, our simple compiler will handle a few other operators as well.

We will tokenize the input (i.e., write a scanner), model the syntax of the source, and let this syntax direct the translation all the way to three-address code, our intermediate language.

2.2: Syntax Definition

2.2.1: Definition of Grammars

This will be “done right” in the next two chapters.

A context-free grammar (CFG) consists of

  1. A set of terminal tokens.
  2. A set of nonterminals.
  3. A set of productions (rules for transforming nonterminals).
  4. A specific nonterminal designated as start symbol.

Example:

    Terminals: 0 1 2 3 4 5 6 7 8 9 + -
    Nonterminals: list digit
    Productions: list → list + digit
                 list → list - digit
                 list → digit
                 digit → 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
    Start symbol: list
  

If no start symbol is specifically designated, the LHS of the first production is the start symbol.

2.2.2: Derivations

Watch how we can generate the input 7+4-5 starting with the start symbol, applying productions, and stopping when no productions are possible (we have only terminals).

    list → list - digit
         → list - 5
         → list + digit - 5
         → list + 4 - 5
         → digit + 4 - 5
         → 7 + 4 - 5
  

This process of applying productions, starting with the start symbol and ending when only terminals are present is called a derivation and we say that the final string has been derived from the initial string (in this case the start symbol).

The set of all strings derivable from the start symbol is the language generated by the CFG

Given a grammar, parsing a string consists of determining if the string is in the language generated by the grammar. If it is in the language, parsing produces a derivation. If it is not, parsing reports an error.

The opposite of derivation is reduction, that is, the LHS of a production, produces the RHS (a derivation) and the RHS is reduced by the production to the LHS.

Homework: 1a, 1c, 2a-c (don't worry about justifying your answers).

2.2.3: Parse trees

parse-tree-2.2

While deriving 7+4-5, one could produce the Parse Tree shown on the right.

You can read off the productions from the tree. For any internal (i.e., non-leaf) tree node, its children give the right hand side (RHS) of a production having the node itself as the LHS.

The leaves of the tree, read from left to right, is called the yield of the tree. We call the tree a derivation of its yield from its root. The tree on the right is a derivation of 7+4-5 from list.

Homework: 1b

2.2.4: Ambiguity

An ambiguous grammar is one in which there are two or more parse trees yielding the same final string. We wish to avoid such grammars.

The grammar above is not ambiguous. For example 1+2+3 can be parsed only one way; the arithmetic must be done left to right. Note that I am not giving a rule of arithmetic, just of this grammar. If you reduced 2+3 to list you would be stuck since it is impossible to further reduce 1+list (said another way it is not possible to derive 1+list from the start symbol).

Homework: 3 (applied only to parts a, b, and c of 2)

2.2.5: Associativity of operators

Our grammar gives left associativity. That is, if you traverse the parse tree in postorder and perform the indicated arithmetic you will evaluate the string left to right. Thus 8-8-8 would evaluate to -8. If you wished to generate right associativity (normally exponentiation is right associative, so 2**3**2 gives 512 not 64), you would change the first two productions to

  list → digit + list
  list → digit - list

Produce in class the parse tree for 7+4-5 with this new grammar.

2.2.6: Precedence of operators

We normally want * to have higher precedence than +. We do this by using an additional nonterminal to indicate the items that have been multiplied. The example below gives the four basic arithmetic operations their normal precedence unless overridden by parentheses. Redundant parentheses are permitted. Equal precedence operations are performed left to right.

  expr   → expr + term | expr - term | term
  term   → term * factor | term / factor | factor
  factor → digit | ( expr )
  digit  → 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9

We use | to indicate that a nonterminal has multiple possible right hand side. So

  A → B | C
is simply shorthand for
  A → B
  A → C

Do the examples 1+2/3-4*5 and (1+2)/3-4*5 on the board.

Note how the precedence is enforced by the grammar; slick!

Statements

Keywords are very helpful for distinguishing statements from one another.

stmt → id := expr
     | if expr then stmt
     | if expr then stmt else stmt
     | while expr do stmt
     | begin opt-stmts end
opt-stmtsstmt-list | ε
stmt-liststmt-list ; stmt | stmt

Remarks:

  1. opt-stmts stands for optional statements. The begin-end block can be empty in some languages.
  2. The ε (epsilon) stands for the empty string.
  3. The use of epsilon productions will add complications.
  4. Some languages do not permit empty blocks For example, Ada has a null statement, which does nothing when executed, for this purpose.
  5. The above grammar is ambiguous!
  6. The notorious “dangling else” problem.
  7. How do you parse if x then if y then z=1 else z=2?

2.3: Syntax-Directed Translation

Specifying the translation of a source language construct in terms of attributes of its syntactic components. The basic idea is use the productions to specify a (typically recursive) procedure for translation. For example, consider the production

    stmt-list → stmt-list ; stmt
  
To process the left stmt-list, we
  1. Call ourselves recursively to process the right stmt-list (which is smaller). This will, say, generate code for all the statements in the right stmt-list.
  2. Call the procedure for stmt, generating code for stmt.
  3. Process the left stmt-list by combining the results for the first two steps as well as what is needed for the semicolon (a terminal so we do not further delegate its actions). In this case we probably concatenate the code for the right stmt-list and stmt.

To avoid having to say the right stmt-list we write the production as

    stmt-list → stmt-list1 ; stmt
  
where the subscript is used to distinguish the two instances of stmt-list.

2.3.1: Postfix notation

attributed-tree

This notation is called postfix because the rule is operator after operand(s). Parentheses are not needed. The notation we normally use is called infix. If you start with an infix expression, the following algorithm will give you the equivalent postfix expression.

One question is, given say 1+2-3, what is E, F and op? Does E=1+2, F=3, and op=+? Or does E=1, F=2-3 and op=+? This is the issue of precedence mentioned above. To simplify the present discussion we will start with fully parenthesized infix expressions.

Example: 1+2/3-4*5

  1. Start with 1+2/3-4*5
  2. Parenthesize (using standard precedence) to get (1+(2/3))-(4*5)
  3. Apply the above rules to calculate P{(1+(2/3))-(4*5)}, where P{X} means “convert the infix expression X to postfix”.
    1. P{(1+(2/3))-(4*5)}
    2. P{(1+(2/3))} P{(4*5)} -
    3. P{1+(2/3)} P{4*5} -
    4. P{1} P{2/3} + P{4} P{5} * -
    5. 1 P{2} P{3} / + 4 5 * -
    6. 1 2 3 / + 4 5 * -

Example: Now do (1+2)/3-4*5

  1. Parenthesize to get ((1+2)/3)-(4*5)
  2. Calculate P{((1+2)/3)-(4*5)}
    1. P{((1+2)/3) P{(4*5)} -
    2. P{(1+2)/3} P{4*5) -
    3. P{(1+2)} P{3} / P{4} P{5} * -
    4. P{1+2} 3 / 4 5 * -
    5. P{1} P{2} + 3 / 4 5 * -
    6. 1 2 + 3 / 4 5 * -

2.3.2: Synthesized Attributes

We want to decorate the parse trees we construct with annotations that give the value of certain attributes of the corresponding node of the tree. We will do the example of translating infix to postfix with 1+2/3-4*5. We use the following grammar, which follows the normal arithmetic terminology where one multiplies and divides factors to obtain terms, which in turn are added and subtracted to form expressions.

  expr   → expr + term | expr - term | term
  term   → term * factor | term / factor | factor
  factor → digit | ( expr )
  digit  → 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9

This grammar supports parentheses, although our example does not use them. On the right is a movie in which the parse tree is build from this example.

The attribute we will associate with the nodes is the postfix form of the string in the leaves below the node. In particular, the value of this attribute at the root is the postfix form of the entire source.

The book does a simpler grammar (no *, /, or parentheses) for a simpler example. You might find that one easier.

Syntax-Directed Definitions (SDDs)

Definition: A syntax-directed definition is a grammar together with semantic rules associated with the productions. These rules are used to compute attribute values. A parse tree augmented with the attribute values at each node is called an annotated parse tree.

For the bottom-up approach I will illustrate now, we annotate a node after having annotated its children. Thus the attribute values at a node can depend on the children of the node but not the parent of the node. We call these synthesized attributes, since they are formed by synthesizing the attributes of the children.

In chapter 5, when we study top-down annotations as well, we will introduce inherited attributes that are passed down from parents to children.

We specify how to synthesize attributes by giving the semantic rules together with the grammar. That is we give the syntax directed definition.

ProductionSemantic Rule
expr → expr1 + termexpr.t := expr1.t || term.t || '+'
expr → expr1 - termexpr.t := expr1.t || term.t || '-'
expr → termexpr.t := term.t
term → term1 * factorterm.t := term1.t || factor.t || '*'
term → term1 / factorterm.t := term1.t || factor.t || '/'
term → factorterm.t := factor.t
factor → digitfactor.t := digit.t
factor → ( expr )factor.t := expr.t
digit → 0digit.t := '0'
digit → 1digit.t := '1'
digit → 2digit.t := '2'
digit → 3digit.t := '3'
digit → 4digit.t := '4'
digit → 5digit.t := '5'
digit → 6digit.t := '6'
digit → 7digit.t := '7'
digit → 8digit.t := '8'
digit → 9digit.t := '9'

We apply these rules bottom-up (starting with the geographically lowest productions, i.e., the lowest lines on the page) and get the annotated graph shown on the right. The annotation are drawn in green.

Homework: Draw the annotated graph for (1+2)/3-4*5.

2.3.3: Simple Syntax-Directed Definitions

If the semantic rules of a syntax-directed definition all have the property that the new annotation for the left hand side (LHS) of the production is just the concatenation of the annotations for the nonterminals on the RHS in the same order as the nonterminals appear in the production, we call the syntax-directed definition simple. It is still called simple if new strings are interleaved with the original annotations. So the example just done is a simple syntax-directed definition.

Remark: SDD's feature semantic rules. We will soon learn about Translation Schemes, which feature a related concept called semantic actions. When one has a simple SDD, the corresponding translation scheme can be done without constructing the parse tree. That is, while doing the parse, when you get to the point where you would construct the node, you just do the actions. In the corresponding translation scheme for present example, the action at a node is just to print the new strings at the appropriate points.

2.3.4: Tree Traversals

When traversing a tree, there are several choices as to when to visit a given node. The traversal can visit the node

  1. Before visiting any children.
  2. Between visiting children.
  3. After visiting all the children

I do not like the book's code as I feel the names chosen confuses the traversal with visiting the nodes. I prefer the following pseudocode, which also illustrates traversals that are not depth first. Comments are introduced by -- and terminate at the end of the line.

    procedure traverse (n: node)
        -- visit(n);   before children
        if (n is a leaf)  return;
        c = first child;
        traverse (c);
        while more children
             -- visit (n);   between children
             c = next child;
             traverse (c);
        end while;
        -- visit (n);  after children
        end traverse;
  
Lab 1 assigned. See the home page.

In general, SDDs do not impose an evaluation order for the attributes of the parse tree. The only requirement is that each attribute is evaluated after all those that it depends on. This general case is quite difficult and sometimes no such order is possible. Since, at this point in the course, we are considering only synthesized attributes, a depth-first (postorder) traversal will always yield a correct evaluation order for the attributes. This is so since synthesized attributes depend only on attributes of child nodes and a depth-first (postorder) traversal visits a node only after all the children have been visited (and hence all the child node attributes have been evaluated).

2.3.5: Translation schemes

The bottom-up annotation scheme just described generates the final result as the annotation of the root. In our infix → postfix example we get the result desired by printing the root annotation. Now we consider another technique that produces its results incrementally.

Instead of giving semantic rules for each production (and thereby generating annotations) we can embed program fragments called semantic actions within the productions themselves.

When drawn in diagrams (e.g., see the diagram below), the semantic action is connected to its node with a distinctive, often dotted, line. The placement of the actions determine the order they are performed. Specifically, one executes the actions in the order they are encountered in a postorder traversal of the tree.

Definition: A syntax-directed translation scheme is a context-free grammar with embedded semantic actions.

For our infix → postfix translator, the parent either just passes on the attribute of its (only) child or concatenates them left to right and adds something at the end. The equivalent semantic actions would be to either print nothing or print the new item.
semantic-action-tree

Emitting a translation

Here are the semantic actions corresponding to a few of the rows of the table above. Note that the actions are enclosed in {}.

    expr → expr + term    { print('+') }
    expr → expr - term    { print('-') }
    term → term / factor  { print('/') }
    term → factor         { null }
    digit → 3             { print('3') }
  

The diagram for 1+2/3-4*5 with attached semantic actions is shown on the right.

Given an input, e.g. our favorite 1+2/3-4*5, we just do a depth first (postorder) traversal of the corresponding diagram and perform the semantic actions as they occur. When these actions are print statements as above, we can be said to be emitting the translation.

Do a depth first traversal of the diagram on the board, performing the semantic actions as they occur, and confirm that the translation emitted is in fact 123/+45*-, the postfix version of 1+2/3-4*5

Homework: Produce the corresponding diagram for (1+2)/3-4*5.

Prefix to infix translation

When we produced postfix, all the prints came at the end (so that the children were already printed. The { actions } do not need to come at the end. We illustrate this by producing infix arithmetic (ordinary) notation from a prefix source.

pre-infix In prefix notation the operator comes first so +1-23 evaluates to zero and +-123 evaluates to 2. Consider the following grammar, which generates the simple language of prefix expressions consisting of addition and subtraction of digits between 1 and 3 without parentheses (prefix notation and postfix notation do not use parentheses).

  P → + P P | - P P | 1 | 2 | 3

The table below shows both the semantic actions and rules used by the translator. Normally, one does not use both actions and rules.

The resulting parse tree for +1-23 with the semantic actions attached is shown on the right. Note that the output language (infix notation) has parentheses.

Prefix to infix translator
Production with Semantic ActionSemantic Rule


P → + { print('(') } P1 { print(')+(') } P2 { print(')') } P.t := '(' || P1.t || ')+(' || P.t || ')'


P → - { print('(') } P1 { print(')-(') } P2 { print(')') } P.t := '(' || P1.t || ')-(' || P.t || ')'


P → 1 { print('1') }P.t := '1'


P → 2 { print('2') }P.t := '2'


P → 3 { print('3') }P.t := '3'

Homework: 2.

2.4: Parsing

Objective: Given a string of tokens and a grammar, produce a parse tree yielding that string (or at least determine if such a tree exists).

We will learn both top-down (begin with the start symbol, i.e. the root of the tree) and bottom up (begin with the leaves) techniques.

In the remainder of this chapter we just do top down, which is easier to implement by hand, but is less general. Chapter 4 covers both approaches.

Tools (so called “parser generators”) often use bottom-up techniques.

In this section we assume that the lexical analyzer has already scanned the source input and converted it into a sequence of tokens. predictive-parsing

2.4.1: Top-down parsing

Consider the following simple language, which derives a subset of the types found in the (now somewhat dated) programming language Pascal. I do not assume you know pascal.

We have two nonterminals, type, which is the start symbol, and simple, which represents the simple types.

There are 8 terminals, which are tokens produced by the lexer and correspond closely with constructs in pascal itself. Specifically, we have.

  1. integer and char
  2. id for identifier
  3. array and of used in array declarations
  4. ↑ meaning pointer to
  5. num for a (positive whole) number
  6. dotdot for .. (used to give a range like 6..9)

The productions are

    type →   simple
    type →   ↑ id
    type →   array [ simple ] of type
    simple → integer
    simple → char
    simple → num dotdot num
  

Parsing is easy in principle and for certain grammars (e.g., the one above) it actually is easy. We start at the root since this is top-down parsing and apply the two fundamental steps.

  1. At the current (nonterminal) node, select a production whose LHS is this nonterminal and whose RHS matches the input at this point. Make the RHS the children of this node (one child per RHS symbol).
  2. Go to the next node needing a subtree.

When programmed this becomes a procedure for each nonterminal that chooses a production for the node and calls procedures for each nonterminal in the RHS. Thus it is recursive in nature and descends the parse tree. We call these parsers recursive descent.

The big problem is what to do if the current node is the LHS of more than one production. The small problem is what do we mean by the next node needing a subtree.

The easiest solution to the big problem would be to assume that there is only one production having a given terminal as LHS. There are two possibilities

  1. No circularity. For example
        expr → term + term - 9
        term → factor / factor
        factor → digit
        digit → 7
      
    But this is very boring. The only possible sentence is 7/7+7/7-9

  2. Circularity
        expr → term + term
        term → factor / factor
        factor → ( expr )
      
    This is even worse; there are no (finite) sentences. Only an infinite sentence beginning (((((((((.

So this won't work. We need to have multiple productions with the same LHS.

How about trying them all? We could do this! If we get stuck where the current tree cannot match the input we are trying to parse, we would backtrack.

Instead, we will look ahead one token in the input and only choose productions that can yield a result starting with this token. Furthermore, we will (in this section) restrict ourselves to predictive parsing in which there is only production that can yield a result starting with a given token. This solution to the big problem also solves the small problem. Since we are trying to match the next token in the input, we must choose the leftmost (nonterminal) node to give children to.

2.4.2: Predictive parsing

Let's return to pascal array type grammar and consider the three productions having type as LHS. Even when I write the short form
type → simple | ↑ id | array [ simple ] of type
I view it as three productions.

For each production P we wish to consider the set FIRST(P) consisting of those tokens that can appear as the first symbol of a string derived from the RHS of P. FIRST is actually defined on strings not productions. When I write FIRST(P), I really mean FIRST(RHS). Similarly, I often say the first set of the production P when I should really say the first set of the RHS of the production P.

Definition: Let r be the RHS of a production P. FIRST(r) is the set of tokens that can appear as the first symbol in a string derived from r.

To use predictive parsing, we make the following

Assumption: Let P and Q be two productions with the same LHS. Then FIRST(P) and FIRST(Q) are disjoint. Thus, if we know both the LHS and the token that must be first, there is (at most) one production we can apply. BINGO!

An example of predictive parsing

This table gives the FIRST sets for our pascal array type example.

ProductionFIRST
type → simple{ integer, char, num }
type → ↑ id{ ↑ }
type → array [ simple ] of type{ array }
simple → integer{ integer }
simple → char{ char }
simple → num dotdot num{ num }

The three productions with type as LHS have disjoint FIRST sets. Similarly the three productions with simple as LHS have disjoint FIRST sets. Thus predictive parsing can be used. We process the input left to right and call the current token lookahead since it is how far we are looking ahead in the input to determine the production to use. The movie on the right shows the process in action.

Homework:

A. Construct the corresponding table for

  rest → + term rest | - term rest | term
  term → 1 | 2 | 3
B. Can predictive parsing be used?

End of Homework:.
predictive parsing C

2.4.3: When to Use ε-productions

Not all grammars are as friendly as the last example. The first complication is when ε occurs in a RHS. If this happens or if the RHS can generate ε, then ε is included in FIRST.

But ε would always match the current input position!

The rule is that if lookahead is not in FIRST of any production with the desired LHS, we use the (unique!) production (with that LHS) that has ε as RHS.

Your text does a C instead of a pascal example. The productions are

    stmt → expr ;
         | if ( expr ) stmt
         | for ( optexpr ; optexpr ; optexpr ) stmt
         | other
 optexpr → expr | ε
  

For completeness, on the right is the beginning of a movie for the C example. Note the use of the ε-production at the end since no other entry in FIRST will match ;

Once again, the full story will be revealed in chapter 4 when we do parsing in a more complete manner.

2.4.4: Designing a Predictive Parser

Predictive parsers are fairly easy to construct as we will now see. Since they are recursive descent parsers we go top-down with one procedure for each nonterminal. Do remember that we must have disjoint FIRST sets for all the productions having a given nonterminal as LHS.

  1. For each nonterminal, write a procedure that chooses the unique(!) production having lookahead in its FIRST. Use the ε production if no other production matches. If no production matches and there is no ε production, the parse fails.
  2. These procedures mimic the RHS of the production. They call procedures for each nonterminal and call match for each terminal.
  3. Write a procedure match(terminal) that advances lookahead to the next input token after confirming that the previous value of lookahead equals the terminal argument.
  4. Write a main program that initializes lookahead to the first input token and invokes the procedure for the start symbol.

The book has code at this point, which you should read. We will see code in class, later in this chapter.

2.4.5: Left Recursion

Another complication. Consider
expr → expr + term
expr → term

For the first production the RHS begins with the LHS. This is called left recursion. If a recursive descent parser would pick this production, the result would be that the next node to consider is again expr and the lookahead has not changed. An infinite loop occurs.

Consider instead
expr → term rest
rest → + term rest
rest → ε

Both sets of productions generate the same possible token strings, namely
term + term + ... + term
The second set is called right recursive since the RHS ends (has on the right) the LHS. If you draw the parse trees generated, you will see that, for left recursive productions, the tree grows to the left; whereas, for right recursive, it grows to the right.

Note also that, according to the trees generated by the first pair, the additions are performed right to left; whereas, for the second pair, they are performed left to right. That is, for
term + term + term
the tree from the first pair has the left + at the top (why?); whereas, the tree from the second pair has the right + at the top.

In general, for any A, R, α, and β, we can replace the pair
A → A α | β
with the triple
A → β R
R → α R | ε

For the example above A is expr, R is rest, α is + term, and β is term.

2.5: A Translator for Simple Expressions

Objective: an infix to postfix translator for expressions. We start with just plus and minus, specifically the expressions generated by the following grammar. We include a set of semantic actions with the grammar. Note that finding a grammar for the desired language is one problem, constructing a translator for the language given a grammar is another problem. We are tackling the second problem.

  expr → expr + term { print('+') }
  expr → expr - term { print('-') }
  expr → term
  term → 0           { print('0') }
  . . .
  term → 9           { print('9') }

One problem that we must solve is that this grammar is left recursive.

2.5.1: Abstract and concrete syntax

We prefer not to have superfluous nonterminals as they make the parsing less efficient. That is why we don't say that a term produces a digit and a digit produces each of 0,...,9. Ideally the syntax tree would just have the operators + and - and the 10 digits 0,1,...,9. That would be called the abstract syntax tree. A parse tree coming from a grammar is technically called a concrete syntax tree.

2.5.2: Adapting the Translation Scheme

We eliminate the left recursion as we did in 2.4. This time there are two operators + and - so we replace the triple
A → A α | A β | γ
with the quadruple
A → γ R
R → α R | β R | ε

This time we have actions so, for example
α is + term { print('+') }
However, the formulas still hold and we get

  expr → term rest
  rest → + term { print('+') } rest
       | - term { print('-') } rest
       | ε
  term → 0           { print('0') }
       . . .
       | 9           { print('9') }

2.5.3: Procedures for the nonterminals expr, term, and rest

The C code is in the book. Note the else ; in rest(). This corresponds to the epsilon production. As mentioned previously. The epsilon production is only used when all others fail (that is why it is the else arm and not the then or the else if arms).

2.5.4: Simplifying the translator

These are (useful) programming techniques.

The complete program

The program in Java is in the book.

2.6: Lexical analysis

Converts a sequence of characters (the source) into a sequence of tokens. A lexeme is the sequence of characters comprising a single token. The reason we were able to produce the translator in the previous section without a lexer, is that all the tokens were just one character (that is why we had just single digits).

2.6.1: Removal of White space and comments

These do not become tokens so that the parser need not worry about them.

2.6.2: Reading ahead

Consider distinguishing x<y from x<=y.

After reading the < we must read another character. If it is a y, we have found our token (<). However, we must unread the y so that when asked for the next token we will start at y. If it is never more than one extra character that must be examined, a single char variable would suffice. A more general solution is discussed next chapter (Lexical Analysis).

2.6.3: Constants

This chapter considers only numerical integer constants. They are computed one digit at a time by value=10*value+digit. The parser will therefore receive the token num rather than a sequence of digits. Recall that our previous parsers considered only one digit numbers.

The value of the constant can be considered the attribute of the token named num. Alternatively, the attribute can be a pointer/index into the symbol table entry for the number (or into a numbers table).

2.6.4: Recognizing identifiers and keywords

The C statement
sum = sum + x;
contains 4 tokens. The scanner will convert the input into
id = id + id ; (id standing for identifier).
Although there are three id tokens, the first and second represent the lexeme sum; the third represents x. These must be distinguished. Many language keywords, for example then, are syntactically the same as identifiers. These also must be distinguished. The symbol table will accomplish these tasks. We assume (as do most modern languages) that the keywords are reserved, i.e., cannot be used as program variables. The we simply initialize the symbol table to contain all these reserved words and mark them as keywords. When the lexer encounters a would-be identifier and searches the symbol table, it finds out that the string is actually a keyword.

Care must be taken when one lexeme is a proper subset of another. Consider
x<y versus x<=y
When the < is read, the scanner needs to read another character to see if it is an =. But if that second character is y, the current token is < and the y must be “pushed back” onto the input stream so that the configuration is the same after scanning < as it is after scanning <=.

Also consider then versus thenewvalue, one is a keyword and the other an id.

2.6.5: A lexical analyzer

A Java program is given. The book, but not the course, seems to assume knowledge of Java.

Since the scanner converts digits into num's we can shorten the grammar. Here is the shortened version before the elimination of left recursion. Note that the value attribute of a num is its numerical value.

  expr   → expr + term    { print('+') }
  expr   → expr - term    { print('-') }
  expr   → term
  term   → num            { print(num,value) }
In anticipation of other operators with higher precedence, we could introduce factor and, for good measure, include parentheses for overriding the precedence. Our grammar would then become.
  expr   → expr + term    { print('+') }
  expr   → expr - term    { print('-') }
  expr   → term
  term   → factor
  factor → ( expr ) | num { print(num,value) }

The factor() procedure follows the familiar recursive descent pattern: find a production with lookahead in FIRST and do what the RHS says.

2.7: Incorporating a symbol table

The symbol table is an important data structure for the entire compiler. One example of its use would be for semantic actions associated with declarations to set the type field of an entry; semantic actions associated with expression evaluation would used this type information. For the simple infix to postfix translator (which is typeless), the table is primarily used to store and retrieve <lexeme,token> pairs.

2.7.1: Symbol Table per Scope

There is a serious issue here involving scope. We will learn soon that lexers are based on regular expressions; whereas parsers are based on the stronger but more expensive context-free grammars. Regular expressions are not powerful enough to handle nested scopes. So, if the language you are compiling supports nested scopes, the lexer can only construct the <lexeme,token> pairs. The parser converts these pairs into a true symbol table that reflects the nested scopes. If the language is flat, the scanner can produce the symbol table.

The idea is that, when entering a block, a new symbol table is created. Each such table points to the one immediately outer. This structure supports the most-closely nested rule for symbols: a symbol is in the scope of most-closely nested declaration. This gives rise to a tree of tables.

Interface

Reserved keywords

Simply insert them into the symbol table prior to examining any input. Then they can be found when used correctly and, since their corresponding token will not be id, any use of them where an identifier is required can be flagged. For example one would have insert(int) performed for every table.

2.7.2: The Use of Symbol Tables

Below is the grammar for a stripped down example showing nested scopes. The language consists just of declarations of the form
identifier : type ; -- I like ada not C style declarations
trivial statements of the form
identifier ;
and nested blocks.

  program → block
  block   → { decls stmts }     -- { } are terminals not actions
  decls   →  decls decl | ε     -- study this one
  decl    → id : type ;
  stmts   → stmts stmt | ε      -- same idea, a list
  stmt    → block | factor ;      -- get nested block
  factor  → id
Semantic Actions
ProductionAction




Program{top = null}
block




block{{ saved = top;
   top = new Env(top);
   print ("} "); }
decls stmts }{ top = saved;
   print ("} "); }




declsdecls decl
| ε




decltype id ;{ s = new Symbol;
   s.type = type.lexeme;
   top.put(id.lexeme,s); }




stmtsstmts stmt
| ε




stmtblock
| factor ; { print("; "); }




factorid{ s = top.get(id.lexeme);
   print(s.type); }

One possible program in this language is

    { x : int ;   y : float ;
      x ;   y ;
      {  x : float ;
         x ;   y ;
      }
      {  y : int ;
         x ;   y;
      }
      x ;   y ;
    }
  

To show that we have correctly parsed the input and obtained its meaning (i.e., performed semantic analysis) we want to digest the declarations and translate the statements so that we get

  { int ; float ;   { float ; float ; }   { int ; int ; } }
  

The translation scheme, slightly modified from the book page 90, is shown on the right. First a formatting comment.

This translation looks weird, but is actually a good idea (of the authors): it reconciles the two goals of respecting the ordering and nonetheless having the actions all in one column.

Recall that the placement of the actions within the RHS of the production is significant. The RHS is process in order (from left to right) but with a postorder traversal. Thus an action is executed after all the subtrees rooted by parts of the RHS to the left of the action and is executed before all the subtrees rooted by parts of the RHS to the right of the action.

Consider the first production. We want the action to be executed before processing block. Thus the action must precede block in the RHS. But we want the actions in the right column. So we split the RHS over several lines and place an action in the rightmost column of the line that puts in the right order.

The second production has some semantic actions to be performed at the start of the block, and others to be performed at the bottom.

To fully understand the details, you must read the book; but we can see how it works. A new Env initializes a new symbol table; top.put inserts into the symbol table in the environment top; top.get retrieves from that symbol table.

In some sense the star of the show is the simple production
factor → id
together with its semantic actions. These actions look up the identifier in the (correct!) symbol table and print out the type name.

2.8: Intermediate Code Generation

2.8.1: Two kinds of Intermediate Representations

There are two important forms of intermediate representations.

Since parse trees exhibit the syntax of the language being parsed, it may be surprising to see them compared with syntax trees. In fact there is a spectrum of syntax trees, with parse trees within the class.

Another (but less common) name for parse trees is concrete syntax trees. Similarly another (also less common) name for syntax trees is abstract syntax trees.

Very roughly speaking, (abstract) syntax trees are parse trees reduced to their essential components, and three address code looks like assembler without the concept of registers.

2.8.2: Construction of (Abstract) Syntax Trees

Remarks:

  1. Despite the words below, your future lab assignments will likely not require producing abstract syntax trees. Instead, you will be producing concrete syntax trees (parse trees). I will probably include an extra-credit part of some labs that will ask for abstract syntax trees.
  2. Note however that real compilers do not produce parse trees since such trees are larger and have no extra information that the compiler needs. If they produce trees (many do) the produce abstract syntax trees.
  3. The reason I will not require your labs to produce the smaller trees is that to do so it is helpful to understand semantic rules and semantic actions, which come later in the course. Of course, authors of real compilers have already completed the course before starting so this consideration does not apply to them. :-)

Consider the production

    while-stmt → while ( expr ) stmt ;
  
The parse tree would have a node while-stmt with 6 children: while, (, expr, ), stmt, and ;. Many of these are simply syntactic constructs with no real meaning. The essence of the while statement is that the system repeatedly executes stmt until expr is false. Thus, the (abstract) syntax tree has a node (most likely labeled while) with two children, the syntax trees for expr and stmt.

For this while node, we execute

    new While(x,y)
  
where x and y are the already constructed (synthesized attributes!) nodes for expr and stmt

Syntax Trees for Statements

The book has a translation scheme (p.94) for several statements. The part for while reads
  stmt → while ( expr ) stmt1   { stmt.n = new While(expr.n, stmt1.n); }

Representing Blocks in Syntax Trees

Fairly easy

    stmt  → block          { stmt.n = block.n }
    block → { stmts }      { block.n = stmts.n }
  
Together these two just use the syntax tree for the statements constituting the block as the syntax tree for the block when it is used as a statement. So
    while ( x == 5 ) {
       blah
       blah
       more
    }
  
would give the while node of the abstract syntax tree two children as always:
  1. The tree for x==5.
  2. The tree for blah blah more.

Syntax trees for Expressions

When parsing we need to distinguish between + and * to insure that 3+4*5 is parsed correctly, reflecting the higher precedence of *. However, once parsed, the precedence is reflected in the tree itself (the node for + has the node for * as a child). The rest of the compiler treats + and * largely the same so it is common to use the same node label, say OP, for both of them. So we see
    term → term1 * factor     { term.n = new Op('*', term1.n, factor.n); }

2.8.3: Static Checking

Static checking refers to checks performed during compilation; whereas, dynamic checking refers to those performed at run time. Examples of static checks include

L-values and R-values

Consider Q = Z; or A[f(x)+B*D] = g(B+C*h(x,y));. I am using [] for array reference and () for function call).

From a macroscopic view, we have three tasks.

  1. Evaluate the left hand side (LHS) to obtain an l-value.
  2. Evaluate the RHS to obtain an r-value.
  3. Perform the assignment.

Note the differences between L-values, quantities that can appear on the LHS of an assignment, and and R-values, quantities that can appear only on the RHS.

Static checking is used to insure that R-values do not appear on the LHS.

Type Checking

These checks assure that the type of the operands are expected by the operator. In addition to flagging errors, this activity includes

2.8.4: Three-Address Code

These are primitive instructions that have one operator and (up to) three operands, all of which are addresses. One address is the destination, which receives the result of the operation; the other two addresses are the sources of the values to be operated on.

Perhaps the clearest way to illustrate the (up to) three address nature of the instructions is to write them as quadruples or quads.

    ADD        x y z
    MULT       a b c
    ARRAY_L    q r s
    ARRAY_R    e f g
    ifTrueGoto x L
    COPY       r s
  
But we normally write them in a more familiar form.
    x = y + z
    a = b * c
    q[r] = s
    e = f[g]
    ifTrue x goto L
    r = s
  

Translating Statements

We do this and the next section much slower and in much more detail later in the course.

Here is the example from the book, somewhat Java intensive.

    class If extends Stmt {
       Expr E; Stmt S;
       public If(Expr x, Stmt y) { E = x;  S = y;  after = newlabel(); }
       public void gen() {
          Expr n = E.rvalue();
          emit ("ifFalse" + n.toString() + "goto " + after);
          S.gen();
          emit(after + ":");
       }
    }
  

Translating Expressions

I am just illustrating the simplest case

    Expr rvalue(x : Expr) {
       if (x is an Id or Constant node) return x;
       else if (x is an Op(op, y, z) node) {
	  t = new temporary;
	  emit string for t = rvalue(y) op rvalue(z);
	  return a new node for t;
       else read book for other cases
  

Better Code for Expressions

So called optimization (the result is far from optimal) is a huge subject that we barely touch. Here are a few very simple examples. We will cover these since they are local optimizations, that is they occur within a single basic block (a sequence of statements that execute without any jumps).

  1. For a Java assignment statement x = x + 1; we would generate two three-address instructions
    	temp = x + 1
    	x    = temp
          
    The can be combined into the three-address instruction x = x + 1, providing there are no further uses of the temporary.
  2. Common subexpressions occurring in two different expressions, need be computed only once.

Two Questions

  1. How come this compiler was so easy?
  2. Why isn't the final exam next week?

One reason is that much was deliberately simplified. Specifically note that

Also, I presented the material way too fast to expect full understanding.

Chapter 3: Lexical Analysis

Homework: Read chapter 3.

Two methods to construct a scanner (lexical analyzer).

  1. By hand, beginning with a diagram of what lexemes look like. Then write code to follow the diagram and return the corresponding token and possibly other information.
  2. Feed the patterns describing the lexemes to a “lexer-generator”, which then produces the scanner. The historical lexer-generator is Lex; a more modern one is flex.

Note that the speed (of the lexer not of the code generated by the compiler) and error reporting/correction are typically much better for a handwritten lexer. As a result most production-level compiler projects write their own lexers

3.1: the role of the Lexical Analyzer

The lexer is called by the parser when the latter is ready to process another token.

The lexer also might do some housekeeping such as eliminating whitespace and comments. Some call these tasks scanning, but others call the entire task scanning.

After the lexer, individual characters are no longer examined by the compiler; instead tokens (the output of the lexer) are used.

3.1.1: Lexical Analysis Versus Parsing

Why separate lexical analysis from parsing? The reasons are basically software engineering concerns.

  1. Simplicity of design. When one detects a well defined subtask (produce the next token), it is often good to separate out the task (modularity).
  2. Efficiency. With the task separated it is easier to apply specialized techniques.
  3. Portability. Only the lexer need communicate with the outside.

3.1.2: Tokens, Patterns, and Lexemes

Note the circularity of the definitions for lexeme and pattern.

Common token classes.

  1. One for each keyword. The pattern is trivial.
  2. One for each operator or class of operators. A typical class is the comparison operators. Note that these have the same precedence. We might have + and - as the same token, but not + and *.
  3. One for all identifiers (e.g. variables, user defined type names, etc).
  4. Constants (i.e., manifest constants) such as 6 or “hello”, but not a constant identifier such as “quantum” in the Java statement.
    “static final int quantum = 3;”. There might be one token for integer constants, one for real, one for string, etc.
  5. One for each punctuation symbol.

Homework: 3.3.

3.1.3: Attributes for Tokens

We saw an example of attributes in the last chapter.

For tokens corresponding to keywords, attributes are not needed since the name of the token tells everything. But consider the token corresponding to integer constants. Just knowing that the we have a constant is not enough, subsequent stages of the compiler need to know the value of the constant. Similarly for the token identifier we need to distinguish one identifier from another. The normal method is for the attribute to specify the symbol table entry for this identifier.

3.1.4: Lexical Errors

We saw in this movie an example where parsing got “stuck” because we reduced the wrong part of the input string. We also learned about FIRST sets that enabled us to determine which production to apply when we are operating left to right on the input. For predictive parsers the FIRST sets for a given nonterminal are disjoint and so we know which production to apply. In general the FIRST sets might not be disjoint so we have to try all the productions whose FIRST set contains the lookahead symbol.

All the above assumed that the input was error free, i.e. that the source was a sentence in the language. What should we do when the input is erroneous and we get to a point where no production can be applied?

The simplest solution is to abort the compilation stating that the program is wrong, perhaps giving the line number and location where the parser could not proceed.

We would like to do better and at least find other errors. We could perhaps skip input up to a point where we can begin anew (e.g. after a statement ending semicolon), or perhaps make a small change to the input around lookahead so that we can proceed.

3.2: Input Buffering

Determining the next lexeme often requires reading the input beyond the end of that lexeme. For example, to determine the end of an identifier normally requires reading the first whitespace character after it. Also just reading > does not determine the lexeme as it could also be >=. When you determine the current lexeme, the characters you read beyond it may need to be read again to determine the next lexeme.

3.2.1: Buffer Pairs

The book illustrates the standard programming technique of using two (sizable) buffers to solve this problem.

3.2.2: Sentinels

A useful programming improvement to combine testing for the end of a buffer with determining the character read.

3.3: Specification of Tokens

The chapter turns formal and, in some sense, the course begins. The book is fairly careful about finite vs infinite sets and also uses (without a definition!) the notion of a countable set. (A countable set is either a finite set or one whose elements can be put into one to one correspondence with the positive integers. That is, it is a set whose elements can be counted. The set of rational numbers, i.e., fractions in lowest terms, is countable; the set of real numbers is uncountable, because it is strictly bigger, i.e., it cannot be counted.) We should be careful to distinguish the empty set φ from the empty string ε. Formal language theory is a beautiful subject, but I shall suppress my urge to do it right and try to go easy on the formalism.

3.3.1: Strings and Languages

We will need a bunch of definitions.

Definition: An alphabet is a finite set of symbols.

Example: {0,1}, presumably φ (uninteresting), ascii, unicode, ebcdic, latin-1.

Definition: A string over an alphabet is a finite sequence of symbols from that alphabet. Strings are often called words or sentences.

Example: Strings over {0,1}: ε, 0, 1, 111010. Strings over ascii: ε, sysy, the string consisting of 3 blanks.

Definition: The length of a string is the number of symbols (counting duplicates) in the string.

Example: The length of allan, written |allan|, is 5.

Definition: A language over an alphabet is a countable set of strings over the alphabet.

Example: All grammatical English sentences with five, eight, or twelve words is a language over ascii. It is also a language over unicode.

Definition: The concatenation of strings s and t is the string formed by appending the string t to s. It is written st.

Example: εs = sε = s for any string s.

We view concatenation as a product (see Monoid in wikipedia http://en.wikipedia.org/wiki/Monoid). It is thus natural to define s0=ε and si+1=sis.

Example: s1=s, s4=ssss.

More string terminology

A prefix of a string is a portion starting from the beginning and a suffix is a portion ending at the end. More formally,

Definitions: A prefix of s is any string obtained from s by removing (possibly zero) characters from the end of s.

A suffix is defined analogously and a substring of s is obtained by deleting a prefix and a suffix.

Example: If s is 123abc, then
(1) s itself and ε are each a prefix, suffix, and a substring.
(2) 12 are 123a are prefixes.
(3) 3abc is a suffix.
(4) 23a is a substring.

Definitions: A proper prefix of s is a prefix of s other than ε and s itself. Similarly, proper suffixes and proper substrings of s do not include ε and s.

Definition: A subsequence of s is formed by deleting (possibly) positions from s. We say positions rather than characters since s may for example contain 5 occurrences of the character Q and we only want to delete a certain 3 of them.

Example: issssii is a subsequence of Mississippi.

Homework: 3.1b, 3.5 (c and e are optional).

3.3.2: Operations on Languages

Definition: The union of L1 and L2 is simply the set-theoretic union, i.e., it consists of all words (strings) in either L1 or L2.

Example: The union of {Grammatical English sentences with one, three, or five words} with {Grammatical English sentences with two or four words} is {Grammatical English sentences with five or fewer words}.

Definition: The concatenation of L1 and L2 is the set of all strings st, where s is a string of L1 and t is a string of L2.

We again view concatenation as a product and write LM for the concatenation of L and M.

Examples:: The concatenation of {a,b,c} and {1,2} is {a1,a2,b1,b2,c1,c2}. The concatenation of {a,b,c} and {1,2,ε} is {a1,a2,b1,b2,c1,c2,a,b,c}.

Definition: As with strings, it is natural to define powers of a language L.
L0={ε}, which is not φ.
Li+1=LiL.

Definition: The (Kleene) closure of L, denoted L* is
L0 ∪ L1 ∪ L2 ...

Definition: The positive closure of L, denoted L+ is
L1 ∪ L2 ...

Example: {0,1,2,3,4,5,6,7,8,9}+ gives all unsigned integers, but with some ugly versions. It has 3, 03, 000003.
{0} ∪ ( {1,2,3,4,5,6,7,8,9} ({0,1,2,3,4,5,6,7,8,9,0}* ) ) seems better.

In these notes I may write * for * and + for +, but that is strictly speaking wrong and I will not do it on the board or on exams or on lab assignments.

Example: {a,b}* is {ε,a,b,aa,ab,ba,bb,aaa,aab,aba,abb,baa,bab,bba,bbb,...}.
{a,b}+ is {a,b,aa,ab,ba,bb,aaa,aab,aba,abb,baa,bab,bba,bbb,...}.
{ε,a,b}* is {ε,a,b,aa,ab,ba,bb,...}.
{ε,a,b}+ is the same as {ε,a,b}*.

The book gives other examples based on L={letters} and D={digits}, which you should read..

3.3.3: Regular Expressions

The idea is that the regular expressions over an alphabet consist of
ε, the alphabet, and expressions using union, concatenation, and *,
but it takes more words to say it right. For example, I didn't include (). Note that (A ∪ B)* is definitely not A* ∪ B* (* does not distribute over ∪) so we need the parentheses.

The book's definition includes many () and is more complicated than I think is necessary. However, it has the crucial advantages of being correct and precise.

The wikipedia entry doesn't seem to be as precise.

I will try a slightly different approach, but note again that there is nothing wrong with the book's approach (which appears in both first and second editions, essentially unchanged).

Definition: The regular expressions and associated languages over an alphabet consist of

  1. ε, the empty string; the associated language L(ε) is {ε}.
  2. Each symbol x in the alphabet; L(x) is {x}.
  3. rs for all regular expressions (REs) r and s; L(rs) is L(r)L(s).
  4. r|s for all REs r and s; L(r|s) is L(r) ∪ L(s).
  5. r* for all REs r; L(r*) is (L(r))*.
  6. (r) for all REs r; L((r)) is L(r).

Parentheses, if present, control the order of operations. Without parentheses the following precedence rules apply.

The postfix unary operator * has the highest precedence. The book mentions that it is left associative. (I don't see how a postfix unary operator can be right associative or how a prefix unary operator such as unary - could be left associative.)

Concatenation has the second highest precedence and is left associative.

| has the lowest precedence and is left associative.

The book gives various algebraic laws (e.g., associativity) concerning these operators.

The reason we don't include the positive closure is that for any RE
r+ = rr*.

Homework: 3.6 a and b.

3.3.4: Regular Definitions

These will look like the productions of a context free grammar we saw previously, but there are differences. Let Σ be an alphabet, then a regular definition is a sequence of definitions

  d1 → r1
  d2 → r2
      ...
  dn → rn
where the d's are unique and not in Σ and
ri is a regular expressions over Σ ∪ {d1,...,di-1}.

Note that each di can depend on all the previous d's.

Example: C identifiers can be described by the following regular definition

  letter_ → A | B | ... | Z | a | b | ... | z | _
    digit → 0 | 1 | ... | 9
      CId → letter_ ( letter_ | digit)*

Homework: 3.7 a,b (c is optional)

3.3.5: Extensions of Regular Expressions

There are many extensions of the basic regular expressions given above. The following three will be frequently used in this course as they are particular useful for lexical analyzers as opposed to text editors or string oriented programming languages, which have more complicated regular expressions.

All three are simply shorthand. That is, the set of possible languages generated using the extensions is the same as the set of possible languages generated without using the extensions.

  1. One or more instances. This is the positive closure operator + mentioned above.
  2. Zero or one instance. The unary postfix operator ? defined by
    r? = r | ε for any RE r.
  3. Character classes. If a1, a2, ..., an are symbols in the alphabet, then
    [a1a2...an] = a1 | a2 | ... | an. In the special case where all the a's are consecutive, we can simply the notation further to just [a1-an].

Examples:

C-language identifiers

  letter_ → [A-Za-z_]
    digit → [0-9]
      CId → letter_ ( letter | digit ) *
  

Unsigned integer or floating point numbers

  digit → [0-9]
 digits → digit+
 number → digits (. digits)?(E[+-]? digits)?

Homework: 3.8 for the C language (you might need to read a C manual first to find out all the numerical constants in C), 3.10a.

3.4: Recognition of Tokens

Goal is to perform the lexical analysis needed for the following grammar.

  stmt → if expr then stmt
       | if expr then stmt else stmt
       | ε
  expr → term relop term      // relop is relational operator =, >, etc
       | term
  term →  id
       | number

Recall that the terminals are the tokens, the nonterminals produce terminals.

A regular definition for the terminals is

  digit → [0-9]
 digits → digits+
 number → digits (. digits)? (E[+-]? digits)?
 letter → [A-Za-z]
     id → letter ( letter | digit )*
     if → if
   then → then
   else → else
  relop → < | > | <= | >= | = | <>
  
LexemeTokenAttribute
Whitespacews
ifif
thenthen
elseelse
An identifieridPointer to table entry
A numbernumberPointer to table entry
<relopLT
<=relopLE
=relopEQ
<>relopNE
>relopGT
>=relopGE

We also want the lexer to remove whitespace so we define a new token

  ws → ( blank | tab | newline ) +
where blank, tab, and newline are symbols used to represent the corresponding ascii characters.

Recall that the lexer will be called by the parser when the latter needs a new token. If the lexer then recognizes the token ws, it does not return it to the parser but instead goes on to recognize the next token, which is then returned. Note that you can't have two consecutive ws tokens in the input because, for a given token, the lexer will match the longest lexeme starting at the current position that yields this token. The table on the right summarizes the situation.

For the parser all the relational ops are to be treated the same so they are all the same token, relop. Naturally, other parts of the compiler will need to distinguish between the various relational ops so that appropriate code is generated. Hence, they have distinct attribute values. trans dia relop

3.4.1: Transition Diagrams

A transition diagram is similar to a flowchart for (a part of) the lexer. We draw one for each possible token. It shows the decisions that must be made based on the input seen. The two main components are circles representing states (think of them as decision points of the lexer) and arrows representing edges (think of them as the decisions made).

The transition diagram (3.12 in the 1st edition, 3.13 in the second) for relop is shown on the right.

  1. The double circles represent accepting or final states at which point a lexeme has been found. There is often an action to be done (e.g., returning the token), which is written to the right of the double circle.
  2. If we have moved one (or more) characters too far in finding the token, one (or more) stars are drawn.
  3. An imaginary start state exists and has an arrow coming from it to indicate where to begin the process.

It is fairly clear how to write code corresponding to this diagram. You look at the first character, if it is <, you look at the next character. If that character is =, you return (relop,LE) to the parser. If instead that character is >, you return (relop,NE). If it is another character, return (relop,LT) and adjust the input buffer so that you will read this character again since you have used it for the current lexeme. If the first character was =, you return (relop,EQ).

3.4.2: Recognition of Reserved Words and Identifiers

The transition diagram below corresponds to the regular definition given previously.

trans dia id

Note again the star affixed to the final state.

Two questions remain.

  1. How do we distinguish between identifiers and keywords such as then, which also match the pattern in the transition diagram?
  2. What is (gettoken(), installID())?

We will continue to assume that the keywords are reserved, i.e., may not be used as identifiers. (What if this is not the case—as in Pl/I, which had no reserved words? Then the lexer does not distinguish between keywords and identifiers and the parser must.)

We will use the method mentioned last chapter and have the keywords installed into the symbol table prior to any invocation of the lexer. The symbol table entry will indicate that the entry is a keyword.

installID() checks if the lexeme is already in the table. If it is not present, the lexeme is install as an id token. In either case a pointer to the entry is returned.

gettoken() examines the lexeme and returns the token name, either id or a name corresponding to a reserved keyword.

Both installID() and gettoken() access the buffer to obtain the lexeme of interest

The text also gives another method to distinguish between identifiers and keywords.

3.4.3: Completion of the Running Example

So far we have transition diagrams for identifiers (this diagram also handles keywords) and the relational operators. What remains are whitespace, and numbers, which are the simplest and most complicated diagrams seen so far.

Recognizing Whitespace

The diagram itself is quite simple reflecting the simplicity of the corresponding regular expression.

trans dia ws

Recognizing Numbers

The diagram below is from the second edition. It is essentially a combination of the three diagrams in the first edition.

trans dia num

This certainly looks formidable, but it is not that bad; it follows from the regular expression.

In class go over the regular expression and show the corresponding parts in the diagram.

When an accepting states is reached, action is required but is not shown on the diagram. Just as identifiers are stored in a symbol table and a pointer is returned, there is a corresponding number table in which numbers are stored. These numbers are needed when code is generated. Depending on the source language, we may wish to indicate in the table whether this is a real or integer. A similar, but more complicated, transition diagram could be produced if they language permitted complex numbers as well.

Homework: Write transition diagrams for the regular expressions in problems 3.6 a and b, 3.7 a and b.

3.4.4: Architecture of a Transition-Diagram-Based Lexical Analyzer

The idea is that we write a piece of code for each decision diagram. I will show the one for relational operations below (from the 2nd edition). This piece of code contains a case for each state, which typically reads a character and then goes to the next case depending on the character read. The numbers in the circles are the names of the cases.

Accepting states often need to take some action and return to the parser. Many of these accepting states (the ones with stars) need to restore one character of input. This is called retract() in the code.

What should the code for a particular diagram do if at one state the character read is not one of those for which a next state has been defined? That is, what if the character read is not the label of any of the outgoing arcs? This means that we have failed to find the token corresponding to this diagram.

The code calls fail(). This is not an error case. It simply means that the current input does not match this particular token. So we need to go to the code section for another diagram after restoring the input pointer so that we start the next diagram at the point where this failing diagram started. If we have tried all the diagram, then we have a real failure and need to print an error message and perhaps try to repair the input.

Note that the order the diagrams are tried is important. If the input matches more than one token, the first one tried will be chosen.

  TOKEN getRelop()                        // TOKEN has two components
    TOKEN retToken = new(RELOP);          // First component set here
    while (true)
       switch(state)
         case 0: c = nextChar();
                 if (c == '<')      state = 1;
                 else if (c == '=') state = 5;
                 else if (c == '>') state = 6;
                 else fail();
                 break;
         case 1: ...
         ...
         case 8: retract();  // an accepting state with a star
                 retToken.attribute = GT;  // second component
                 return(retToken);
  

Second edition additions

The description above corresponds to the one given in the first edition.

The newer edition gives two other methods for combining the multiple transition-diagrams (in addition to the one above).

  1. Unlike the method above, which tries the diagrams one at a time, the first new method tries them in parallel. That is, each character read is passed to each diagram (that hasn't already failed). Care is needed when one diagram has accepted the input, but others still haven't failed and may accept a longer prefix of the input.
  2. The final possibility discussed, which appears to be promising, is to combine all the diagrams into one. That is easy for the example we have been considering because all the diagrams begin with different characters being matched. Hence we just have one large start with multiple outgoing edges. It is more difficult when there is a character that can begin more than one diagram.

3.5: The Lexical Analyzer Generator Lex

The newer version, which we will use, is called flex, the f stands for fast. I checked and both lex and flex are on the cs machines. I will use the name lex for both.

Lex is itself a compiler that is used in the construction of other compilers (its output is the lexer for the other compiler). The lex language, i.e, the input language of the lex compiler, is described in the few sections. The compiler writer uses the lex language to specify the tokens of their language as well as the actions to take at each state.

3.5.1: Use of Lex

Let us pretend I am writing a compiler for a language called pink. I produce a file, call it lex.l, that describes pink in a manner shown below. I then run the lex compiler (a normal program), giving it lex.l as input. The lex compiler output is always a file called lex.yy.c, a program written in C.

One of the procedures in lex.yy.c (call it pinkLex()) is the lexer itself, which reads a character input stream and produces a sequence of tokens. pinkLex() also sets a global value yylval that is shared with the parser. I then compile lex.yy.c together with a the parser (typically the output of lex's cousin yacc, a parser generator) to produce say pinkfront, which is an executable program that is the front end for my pink compiler.

3.5.2: Structure of Lex Programs

The general form of a lex program like lex.l is

  declarations
  %%
  translation rules
  %%
  auxiliary functions

The lex program for the example we have been working with follows (it is typed in straight from the book).

  %{
      /* definitions of manifest constants
         LT, LE, EQ, NE, GT, GE,
         IF, THEN, ELSE, ID, NUMBER, RELOP */
  %}

  /* regular definitions */
  delim     [ \t\n]
  ws        {delim}*
  letter    [A-Za-z]
  digit     [0-9]
  id        {letter}({letter}{digit})*
  number    {digit}+(\.{digit}+)?(E[+-]?{digit}+)?

  %%

  {ws}      {/* no action and no return */}
  if        {return(IF);}
  then      {return(THEN);}
  else      {return(ELSE);}
  {id}      {yylval = (int) installID(); return(ID);}
  {number}  {yylval = (int) installNum(); return(NUMBER);}
  "<"       {yylval = LT; return(RELOP);}
  "<="      {yylval = LE; return(RELOP);}
  "="       {yylval = EQ; return(RELOP);}
  "<>"      {yylval = NE; return(RELOP);}
  ">"       {yylval = GT; return(RELOP);}
  ">="      {yylval = GE; return(RELOP);}

  %%

  int installID() {/* function to install the lexeme, whose first character
                      is pointed to by yytext, and whose length is yyleng,
                      into the symbol table and return a pointer thereto    */
  }

  int installNum() {/* similar to installID, but puts numerical constants
                       into a separate table                              */

The first, declaration, section includes variables and constants as well as the all-important regular definitions that define the building blocks of the target language, i.e., the language that the generated lexer will analyze.

The next, translation rules, section gives the patterns of the lexemes that the lexer will recognize and the actions to be performed upon recognition. Normally, these actions include returning a token name to the parser and often returning other information about the token via the shared variable yylval.

If a return is not specified the lexer continues executing and finds the next lexeme present.

Comments on the Lex Program

Anything between %{ and %} is not processed by lex, but instead is copied directly to lex.yy.c. So we could have had statements like

  #define LT 12
  #define LE 13

The regular definitions are mostly self explanatory. When a definition is later used it is surrounded by {}. A backslash \ is used when a special symbol like * or . is to be used to stand for itself, e.g. if we wanted to match a literal star in the input for multiplication.

Each rule is fairly clear: when a lexeme is matched by the left, pattern, part of the rule, the right, action, part is executed. Note that the value returned is the name (an integer) of the corresponding token. For simple tokens like the one named IF, which correspond to only one lexeme, no further data need be sent to the parser. There are several relational operators so a specification of which lexeme matched RELOP is saved in yylval. For id's and numbers's, the lexeme is stored in a table by the install functions and a pointer to the entry is placed in yylval for future use.

Everything in the auxiliary function section is copied directly to lex.yy.c. Unlike declarations enclosed in %{ %}, however, auxiliary functions may be used in the actions

3.5.3: Conflict Resolution in Lex

  1. Match the longest possible prefix of the input.
  2. If this prefix matches multiple patterns, choose the first.

The first rule makes <= one instead of two lexemes. The second rule makes if a keyword and not an id.

3.5.3a: Anger Management in Lex

Sorry.

3.5.4: The Lookahead Operator

Sometimes a sequence of characters is only considered a certain lexeme if the sequence is followed by specified other sequences. Here is a classic example. Fortran, PL/I, and some other languages do not have reserved words. In Fortran
IF(X)=3
is a legal assignment statement and the IF is an identifier. However,
IF(X.LT.Y)X=Y
is an if/then statement and IF is a keyword. Sometimes the lack of reserved words makes lexical disambiguation impossible, however, in this case the slash / operator of lex is sufficient to distinguish the two cases. Consider

  IF / \(.*\){letter}

This only matches IF when it is followed by a ( some text a ) and a letter. The only FORTRAN statements that match this are the if/then shown above; so we have found a lexeme that matches the if token. However, the lexeme is just the IF and not the rest of the pattern. The slash tells lex to put the rest back into the input and match it for the next and subsequent tokens.

Homework: 3.11.

Homework: Modify the lex program in section 3.5.2 so that: (1) the keyword while is recognized, (2) the comparison operators are those used in the C language, (3) the underscore is permitted as another letter (this problem is easy).

3.6: Finite Automata

The secret weapon used by lex et al to convert (compile) its input into a lexer.

Finite automata are like the graphs we saw in transition diagrams but they simply decide if a sentence (input string) is in the language (generated by our regular expression). That is, they are recognizers of language.

There are two types of finite automata

  1. Deterministic finite automata (DFA) have for each state (circle in the diagram) exactly one edge leading out for each symbol. So if you know the next symbol and the current state, the next state is determined. That is, the execution is deterministic; hence the name.
  2. Nondeterministic finite automata (NFA) are the other kind. There are no restrictions on the edges leaving a state: there can be several with the same symbol as label and some edges can be labeled with ε. Thus there can be several possible next states from a given state and a current lookahead symbol.

Surprising Theorem: Both DFAs and NFAs are capable of recognizing the same languages, the regular languages, i.e., the languages generated by regular expressions (plus the automata can recognize the empty language).

What does this mean?

There are certainly NFAs that are not DFAs. But the language recognized by each such NFA can also be recognized by at least one DFA.

Why mention (confusing) NFAs?

The DFA that recognizes the same language as an NFA might be significantly larger that the NFA.

The finite automaton that one constructs naturally from a regular expression is often an NFA.

3.6.1: Nondeterministic Finite Automata

Here is the formal definition.

A nondeterministic finite automata (NFA) consists of

  1. A finite set of states S.
  2. An input alphabet Σ not containing ε.
  3. A transition function that gives, for each state and each symbol in Σ ∪ ε, a set of next states (or successor states.
  4. An element s0 of S, the start state.
  5. A subset F of S, the accepting states (or final states).

An NFA is basically a flow chart like the transition diagrams we have already seen. Indeed an NFA (or a DFA, to be formally defined soon) can be represented by a transition graph whose nodes are states and whose edges are labeled with elements of Σ ∪ ε. The differences between a transition graph and our previous transition diagrams are:

  1. Possibly multiple edges with the same label leaving a single state.
  2. An edge may be labeled with ε.

nfa-24

The transition graph to the right is an NFA for the regular expression (a|b)*abb, which (given the alphabet {a,b} represents all words ending in abb.

Consider aababb. If you choose the wrong edge for the initial a's you will get stuck or not end at the accepting state. But an NFA accepts a word if any path (beginning at the start state and using the symbols in the word in order) ends at an accepting state. It essentially tries all such paths at once and accepts if any end at an accepting state.

Patterns like (a|b)*abb are useful regular expressions! If the alphabet is ascii, consider *.java.
hw sect 3.6

Homework: For the NFA to the right, indicate all the paths labeled aabb.

3.6.2: Transition Tables

Stateabε
0{0,1}{0}φ
1φ{2}φ
2φ{3}φ

There is an equivalent way to represent an NFA, namely a table giving, for each state s and input symbol x (and ε), the set of successor states x leads to from s. The empty set φ is used when there is no edge labeled x emanating from s. The table on the right corresponds to the transition graph above.

The downside of these tables is their size, especially if most of the entries are φ since those entries would not take any space in a transition graph.

Homework: Construct the transition table for the NFA in the previous homework problem.

3.6.3: Acceptance of Input Strings by Automata

An NFA accepts a string if the symbols of the string specify a path from the start to an accepting state.

Homework: Does the NFA in the previous homework accept the string aabb?

Again note that these symbols may specify several paths, some of which lead to accepting states and some that don't. In such a case the NFA does accept the string; one successful path is enough.

Also note that if an edge is labeled ε, then it can be taken for free.

For the transition graph above any string can just sit at state 0 since every possible symbol (namely a or b) can go from state 0 back to state 0. So every string can lead to a non-accepting state, but that is not important since if just one path with that string leads to an accepting state, the NFA accepts the string.

The language defined by an NFA or the language accepted by an NFA is the set of strings (a.k.a. words) accepted by the NFA.

So the NFA in the diagram above (not the diagram with the homework problem) accepts the same language as the regular expression (a|b)*abb.

If A is an automaton (NFA or DFA) we use L(A) for the language accepted by A.

nfa-26

The diagram on the right illustrates an NFA accepting the language L(aa*|bb*). The path
0 → 3 → 4 → 4 → 4 → 4
shows that bbbb is accepted by the NFA.

Note how the ε that labels the edge 0 → 3 does not appear in the string bbbb since ε is the empty string.

3.6.4: Deterministic Finite Automata

There is something weird about an NFA if viewed as a model of computation. How is a computer of any realistic construction able to check out all the (possibly infinite number of) paths to determine if any terminate at an accepting state?

We now consider a much more realistic model, a DFA.

Definition: A deterministic finite automata or DFA is a special case of an NFA having the restrictions

  1. No edge is labeled with ε
  2. For any state s and symbol a, there is exactly one edge leaving s with label a.

This is realistic. We are at a state and examine the next character in the string, depending on the character we go to exactly one new state. Looks like a switch statement to me.

Minor point: when we write a transition table for a DFA, the entries are elements not sets so there are no {} present.

Simulating a DFA

Indeed a DFA is so reasonable there is an obvious algorithm for simulating it (i.e., reading a string and deciding whether or not it is in the language accepted by the DFA). We present it now.

The second edition has switched to C syntax: = is assignment == is comparison. I am going to change to this notation since I strongly suspect that most of the class is much more familiar with C/C++/java/C# than with algol60/algol68/pascal/ada (the last is my personal favorite). As I revisit past sections of the notes to fix errors, I will change the examples from algol to C usage of =. I realize that this makes the notes incompatible with the edition you have, but hope and believe that this will not cause any serious problems.

  s = s0;   // start state.  NOTE = is assignment
  c = nextChar();      // a priming read
  while (c != eof) {
    s = move(s,c);
    c = nextChar();
  }
  if (s is in F, the set of accepting states) return yes
  else return no

3.7: From Regular Expressions to Automata

3.7.0: Not losing site of the forest due to the trees

This is not from the book.

Do not forget the goal of the chapter is to understand lexical analysis. We saw, when looking at Lex, that regular expressions are a key in this task. So we want to recognize regular expressions (say the ones representing tokens). We are going to see two methods.

  1. Convert the regular expression to an NFA and simulate the NFA.
  2. Convert the regular expression to an NFA, convert the NFA to a DFA, and simulate the DFA.

So we need to learn 4 techniques.

  1. Convert a regular expression to an NFA
  2. Simulate an NFA
  3. Convert an NFA to a DFA
  4. Simulate a DFA.

The list I just gave is in the order the algorithms would be applied—but you would use either 2 or (3 and 4).

The two editions differ in the order the techniques are presented, but neither does it in the order I just gave. Indeed, we just did item #4.

I will follow the order of 2nd ed but give pointers to the first edition where they differ.

================ Start Lecture #4 ================

Remark: If you find a particular homework question challenging, ask on the mailing list and an answer will be produced.

Remark: I forgot to assign homework for section 3.6. I have added one problem spread into three parts. It is not assigned but it is a question I believe you should be able to do.

3.7.1: Converting an NFA to a DFA

(This is item #3 above and is done in section 3.6 in the first edition.)

The book gives a detailed proof; I am just trying to motivate the ideas.

Let N be an NFA, we construct a DFA D that accepts the same strings as N does. Call a state of N an N-state, and call a state of D a D-state. nfa-34

The idea is that D-state corresponds to a set of N-states and hence this is called the subset algorithm. Specifically for each string X of symbols we consider all the N-states that can result when N processes X. This set of N-states is a D-state. Let us consider the transition graph on the right, which is an NFA that accepts strings satisfying the regular expression
(a|b)*abb.

NFA statesDFA stateab
{0,1,2,4,7}D0D1D2
{1,2,3,4,6,7,8}D1D1D3
{1,2,4,5,6,7}D2D1D2
{1,2,4,5,6,7,9}D3D1D4
{1,2,4,5,6,7,10}D4D1D2

The start state of D is the set of N-states that can result when N processes the empty string ε. This is called the ε-closure of the start state s0 of N, and consists of those N-states that can be reached from s0 by following edges labeled with ε. Specifically it is the set {0,1,2,4,7} of N-states. We call this state D0 and enter it in the transition table we are building for D on the right.

Next we want the a-successor of D0, i.e., the D-state that occurs when we start at D0 and move along an edge labeled a. We call this successor D1. Since D0 consists of the N-states corresponding to ε, D1 is the N-states corresponding to εa=a. We compute the a-successor of all the N-states in D0 and then form the ε-closure.

Next we compute the b-successor of D0 the same way and call it D2.

We continue forming a- and b-successors of all the D-states until no new D-states result (there is only a finite number of subsets of all the N-states so this process does indeed stop).

This gives the table on the right. D4 is the only D-accepting state as it is the only D-state containing the (only) N-accepting state 10.

Theoretically, this algorithm is awful since for a set with k elements, there are 2k subsets. Fortunately, normally only a small fraction of the possible subsets occur in practice.

Homework: Convert the NFA from the homework for section 3.6 to a DFA.

3.7.2: Simulating an NFA

Instead of producing the DFA, we can run the subset algorithm as a simulation itself. This is item #2 in my list of techniques

  S = ε-closure(s0);
  c = nextChar();
  while ( c != eof ) {
    S = ε-closure(move(S,c));
    c = nextChar();
  }
  if ( S ∩ F != φ ) return yes;   // F is accepting states
  else return no;

3.7.3: Efficiency of NFA Simulation

Slick implementation. re to nfa

3.7.4: Constructing an NFA from a Regular Expression

I give a pictorial proof by induction. This is item #1 from my list of techniques.

  1. The base cases are the empty regular expression and the regular expression consisting of a single symbol a in the alphabet.
  2. The inductive cases are.
    1. s | t for s and t regular expressions
    2. st for s and t regular expressions
    3. s*
    4. (s), which is trivial since the nfa for s works for (s).

The pictures on the right illustrate the base and inductive cases.

Remarks:

  1. The generated NFA has at most twice as many states as there are operators and operands in the RE. This is important for studying the complexity of the NFA.
  2. The generated NFA has one start and one accepting state. The accepting state has no outgoing arcs and the start state has no incoming arcs.
  3. Note that the diagram for st correctly indicates that the final state of s and the initial state of t are merged. This uses the previous remark that there is only one start and final state.
  4. Except for the accepting state, each state of the generated NFA has either one outgoing arc labeled with a symbol or two outgoing arcs labeled with ε.

Do the NFA for (a|b)*abb and see that we get the same diagram that we had before.

Do the steps in the normal leftmost, innermost order (or draw a normal parse tree and follow it).

Homework: 3.16 a,b,c

3.7.5: Efficiency of String-Processing Algorithms

(This is on page 127 of the first edition.) Skipped.

3.8: Design of a Lexical-Analyzer Generator

How lexer-generators like Lex work.

3.8.1: The structure of the generated analyzer

We have seen simulators for DFAs and NFAs.

The remaining large question is how is the lex input converted into one of these automatons.

Also

  1. Lex permits functions to be passed through to the yy.lex.c file. This is fairly straightforward to implement.
  2. Lex also supports actions that are to be invoked by the simulator when a match occurs. This is also fairly straight forward.
  3. The lookahead operator is not so simple in the general case and is discussed briefly below.

In this section we will use transition graphs, lexer-generators do not draw pictures; instead they use the equivalent transition tables.

Recall that the regular definitions in Lex are mere conveniences that can easily be converted to REs and hence we need only convert REs into an FSA. nfa png

We already know how to convert a single RE into an NFA. But lex input will contain several REs (since it wishes to recognize several different tokens). The solution is to

  1. Produce an NFA for each RE.
  2. Introduce a new start state.
  3. Introduce an ε transition from the new start state to the start of each NFA constructed in step 1.
  4. When one reaches one of the accepting states,they do NOT stop. See below for an explanation.
The result is shown to the right.

At each of the accepting states (one for each NFA in step 1), the simulator executes the actions specified in the lex program for the corresponding pattern.

3.8.2: Pattern Matching Based on NFAs

We use the algorithm for simulating NFAs presented in 3.7.2.

The simulator starts reading characters and calculates the set of states it is at.

At some point the input character does not lead to any state or we have reached the eof. Since we wish to find the longest lexeme matching the pattern we proceed backwards from the current point (where there was no state) until we reach an accepting state (i.e., the set of NFA states, N-states, contains an accepting N-state). Each accepting N-state corresponds to a matched pattern. The lex rule is that if a lexeme matches multiple patterns we choose the pattern listed first in the lex-program.

PatternAction to perform
aAction1
abbAction2
a*b+Action3

Example

Consider the example on the right with three patterns and their associated actions and consider processing the input aaba.
nfa 52

  1. We begin by constructing the three NFAs. To save space, the third NFA is not the one that would be constructed by our algorithm, but is an equivalent smaller one. For example, some unnecessary ε-transitions have been eliminated. If one view the lex executable as a compiler transforming lex source into NFAs, this would be considered an optimization.
  2. We introduce a new start state and ε-transitions as in the previous section.
  3. We start at the ε-closure of the start state, which is {0,1,3,7}.
  4. The first a (remember the input is aaba) takes us to {2,4,7}. This includes an accepting state and indeed we have matched the first patten. However, we do not stop since we may find a longer match.
  5. The next a takes us to {7}.
  6. The b takes us to {8}.
  7. The next a fails since there are no a-transitions out of state 8. So we must back up to before trying the last a.
  8. We are back in {8} and ask if one of these N-states (I know there is only one, but there could be more) is an accepting state.
  9. Indeed state 8 is accepting for third pattern. If there were more than one accepting state in the list, we would choose the one in the earliest listed pattern.
  10. Action3 would now be performed.
dfa 54

3.8.3: DFA's for Lexical Analyzers

We could also convert the NFA to a DFA and simulate that. The resulting DFA is on the right. Note that it shows the same set of states we had as well as others corresponding other possible inputs.

We label the accepting states with the pattern matched. If multiple patterns are matched (because the accepting D-state contains multiple accepting N-states), we use the first pattern listed (assuming we are using lex conventions).

Technical point. For a DFA, there must be a outgoing edge from each D-state for each possible character. In the diagram, when there is no NFA state possible, we do not show the edge. Technically we should show these edges, all of which lead to the same D-state, called the dead state, and corresponds to the empty subset of N-states.

3.8.4: Implementing the Lookahead Operator

This has some tricky points. Recall that this lookahead operator is for when you must look further down the input but the extra characters matched are not part of the lexeme. We write the pattern r1/r2. In the NFA we match r1 then treat the / as an ε and then match s1. It would be fairly easy to describe the situation when the NFA has only ε-transition at the state where r1 is matched. But it is tricky when there are more than one such transition.

3.9: Optimization of DFA-Based Pattern Matchers

Skipped

3.9.1: Important States of an NFA

Skipped

3.9.2: Functions Computed form the Syntax Tree

Skipped

3.9.3: Computing nullable, firstpos, and lastpos

Skipped

3.9.4: Computing followpos

Skipped

Chapter 4: Syntax Analysis

Homework: Read Chapter 4.

4.1: Introduction

4.1.1: The role of the parser

Conceptually, the parser accepts a sequence of tokens and produces a parse tree.

As we saw in the previous chapter the parser calls the lexer to obtain the next token. In practice this might not occur.

  1. The source program might have errors.
  2. Instead of explicitly constructing the parse tree, the actions that the downstream components of the front end would do on the tree can be integrated with the parser and done incrementally on components of the tree.

There are three classes for grammar-based parsers.

  1. universal
  2. top-down
  3. bottom-up

The universal parsers are not used in practice as they are inefficient.

As expected, top-down parsers start from the root of the tree and proceed downward; whereas, bottom-up parsers start from the leaves and proceed upward.

The commonly used top-down and bottom parsers are not universal. That is, there are grammars that cannot be used with them.

The LL and LR parsers are important in practice. Hand written parsers are often LL. Specifically, the predictive parsers we looked at in chapter two are for LL grammars.

The LR grammars form a larger class. Parsers for this class are usually constructed with the aid of automatic tools.

4.1.2: Representative Grammars

Expressions with + and *

    E → E + T | T
    T → T * F | F
    F → ( E ) | id
  

This takes care of precedence, but as we saw before, gives us trouble since it is left-recursive and we did top-down parsing. So we use the following non-left-recursive grammar that generates the same language.

    E  → T E'
    E' → + T E' | ε
    T  → F T'
    T' → * F T' | ε
    F  → ( E ) | id
  

The following ambiguous grammar will be used for illustration, but in general we try to avoid ambiguity. This grammar does not enforce precedence.

    E → E + E | E * E | ( E ) | id
  

4.1.3: Syntax Error Handling

There are different levels of errors.

  1. Lexical errors: For example, spelling.
  2. Syntactic errors: For example missing ; .
  3. Semantic errors: For example wrong number of array indexes.
  4. Logical errors: For example off by one usage of < instead of <=.

4.1.4: Error-Recovery Strategies

The goals are clear, but difficult.

Trivial Approach: No Recovery

Print an error message when parsing cannot continue and then terminate parsing.

Panic-Mode Recovery

The first level improvement. The parser discards input until it encounters a synchronizing token. These tokens are chosen so that the parser can make a fresh beginning. Good examples are ; and }.

Phrase-Level Recovery

Locally replace some prefix of the remaining input by some string. Simple cases are exchanging ; with , and = with ==. Difficulty is when real error occurred long before the error was detected.

Error Productions

Include productions for common errors.

Global Correction

Change the input I to the closest correct input I' and produce the parse tree for I'.

4.2: Context-Free Grammars

4.2.1: Formal Definition

  1. Terminals: The basic components found by the lexer. They are sometimes called token names, i.e., the first component of the token as produced by the lexer.
  2. Nonterminals: Syntactic variables that help define the syntactic structure of the language.
  3. Start Symbol: A start symbol that is the root of the parse tree.
  4. Productions:
    1. Head or left (hand) side or LHS. A single nonterminal.
    2. Body or right (hand) side or RHS. A string of terminals and nonterminals.

4.2.2: Notational Conventions

I don't use these without saying so.

4.2.3: Derivations

This is mostly (very useful) notation.

Assume we have a production A → α. We would then say that A derives α and write
A ⇒ α

We generalize this. If, in addition, β and γ are strings, we say that βAγ derives βαγ and write
βAγ ⇒ βαγ

We generalize further. If x derives y and y derives z, we say x derives z and write
x ⇒* z.

The notation used is ⇒ with a * over it (I don't see it in html). This should be read derives in zero or more steps. Formally,

  1. x ⇒* x, for any string x.
  2. If x ⇒* y and y ⇒ z, then x ⇒* z.

Definition: If S is the start symbol and S ⇒* x, we say x is a sentential form of the grammar.

A sentential form may contain nonterminals and terminals. If it contains only terminals it is a sentence of the grammar and the language generated by a grammar G, written L(G), is the set of sentences.

Definition: A language generated by a (context-free) grammar is called a context free language.

Definition: Two grammars generating the same language are called equivalent.

Examples: Recall the ambiguous grammar above

    E → E + E | E * E | ( E ) | id
  
We see that id + id is a sentence. Indeed it can be derived in two ways from the start symbol E
    E ⇒ E + E ⇒ id + E ⇒ id + id
    E ⇒ E + E ⇒ E + id ⇒ id + id
  

In the first derivation, we replaced the leftmost nonterminal by the body of a production having the nonterminal as head. This is called a leftmost derivation. Similarly the second derivation in which the rightmost nonterminal is replaced is called a rightmost derivation or a canonical derivation.

When one wishes to emphasize that a (one step) derivation is leftmost they write an lm under the ⇒. To emphasize that a (general) derivation is leftmost, one writes an lm under the ⇒*. Similarly one writes rm to indicate that a derivation is rightmost. I won't do this in the notes but will on the board.

Definition: If x can be derived using a leftmost derivation, we call x a left-sentential form. Similarly for right-sentential form.

================ Start Lecture #5 ================

Homework: 4.1 a, c, d

4.2.4: Parse Trees and Derivations

The leaves of a parse tree (or of any other tree), when read left to right, are called the frontier of the tree. For a parse tree we also call them the yield of the tree.

If you are given a derivation starting with a single nonterminal,

    A ⇒ x1 ⇒ x2 ... ⇒ xn
  
it is easy to write a parse tree with A as the root and xn as the leaves. Just do what (the productions contained in) each step of the derivation says. The LHS of each production is a nonterminal in the frontier of the current tree so replace it with the RHS to get the next tree.

Do this for both the leftmost and rightmost derivations of id+id above.

So there can be many derivations that wind up with the same final tree.

But for any parse tree there is a unique leftmost derivation the produces that tree and a unique rightmost derivation that produces the tree. There may be others as well (e.g., sometime choose the leftmost nonterminal to expand; other times choose the rightmost).

Homework: 4.1 b

4.2.5: Ambiguity

Recall that an ambiguous grammar is one for which there is more than one parse tree for a single sentence. Since each parse tree corresponds to exactly one leftmost (or rightmost) derivation, an ambiguous grammar is one for which there is more than one leftmost (or rightmost) derivation of a given sentence.

We know that the grammar

    E → E + E | E * E | ( E ) | id
  
is ambiguous because we have seen (a few lectures ago) two parse trees for
id + id * id
So there must me at least two leftmost derivations. Here they are
    E ⇒ E + E          E ⇒ E * E
      ⇒ id + E           ⇒ E + E * E
      ⇒ id + E * E       ⇒ id + E * E
      ⇒ id + id * E      ⇒ id + id * E
      ⇒ id + id * id     ⇒ id + id * E
  

As we stated before we prefer unambiguous grammars. Failing that, we want disambiguation rules.

4.2.6: Verification

Skipped

4.2.7: Context-Free Grammars Versus Regular Expressions

Alternatively context-free languages vs regular languages.

Given an RE, construct an NFA as in chapter 3.

From that NFA construct a grammar as follows.

  1. Define a nonterminal Ai for each state i.
  2. For a transition from Ai to Aj on input a (or ε), add a production
    Ai → aAj
  3. If i is accepting, add Ai → ε
  4. If i is start, make Ai start.

If you trace an NFA accepting a sentence, it just corresponds to the constructed grammar deriving the same sentence. Similarly, follow a derivation and notice that at any point prior to acceptance there is only one nonterminal; this nonterminal gives the state in the NFA corresponding to this point in the derivation.

The book starts with (a|b)*abb and then uses the short NFA on the left below. Recall that the NFA generated by our construction is the longer one on the right.

nfa 34 nfa 24

The book gives the simple grammar for the short diagram.

Let's be ambitious and try the long diagram

    A0 → A1 | A7
    A1 → A2 | A4
    A2 → a A3
    A3 → A6
    A4 → b A5
    A5 → A6
    A6 → A1 | A7
    A7 → a A8
    A8 → b A9
    A9 → b A10
    A10 → ε
  

Now trace a path in the NFA and see that it is just a derivation. The same is true in reverse (derivation gives path). The key is that at every stage you have only one nonterminal.

Grammars, but not Regular Expressions, Can Count

The grammar

    A → a A b | ε
  
generates all strings of the form anbn, where there are the same number of a's and b's. In a sense the grammar has counted. No RE can generate this language (proof in book).

4.3: Writing a Grammar

4.3.1: Lexical vs Syntactic Analysis

Why have separate lexer and parser?

  1. As stated before, there are software engineering (modular programming) reasons.
  2. The lexer deals with REs / Regular Languages.
    The parser deals with the more powerful Context Free Grammars / Context Free Languages (CFLs).

4.3.2: Eliminating Ambiguity

Recall the ambiguous grammar with the notorious dangling else problem.

    stmt → if expr then stmt
         | if expr then stmt else stmt
         | other
  

This has two leftmost derivations for
if E1 then S1 else if E2 then S2 else S3

Do these on the board. They differ in the beginning.

In this case we can find a non-ambiguous, equivalent grammar.

        stmt → matched-stmt | open-stmt
matched-stmp → if expr then matched-stmt else matched-stmt
	     | other
   open-stmt → if expr then stmt
	     | if expr then matched-stmt else open-stmt
  

On the board try to find leftmost derivations of the problem sentence above.

4.3.3: Eliminating Left Recursion

We did special cases in chapter 2. Now we do it right(tm).

Previously we did it separately for one production and for two productions with the same nonterminal A on the LHS. Not surprisingly, this can be done for n such productions (together with other non-left recursive productions involving A).

Specifically we start with

    A → A x1 | A x2 | ... A xn | y1 | y2 | ... ym
  
where the x's and y's are strings, no x is ε, and no y begins with A.

The equivalent non-left recursive grammar is

    A  → y1 A' | ... | ym A'
    A' → x1 A' | ... | xn A' | ε
  

Example: Assume x1 is + and y1 is *. With the recursive grammar, we have the following lm derivation.
A ⇒ A + ⇒ , +
With the non-recursive grammar we have
A ⇒ , A' ⇒ , + A' ⇒ , +

This removes direct left recursion where a production with A on the left hand side begins with A on the right. If you also had direct left recursion with B, you would apply the procedure twice.

The harder general case is where you permit indirect left recursion, where, for example one production has A as the LHS and begins with B on the RHS, and a second production has B on the LHS and begins with A on the RHS. Thus in two steps we can turn A into something starting again with A. Naturally, this indirection can involve more than 2 nonterminals.

Theorem: All left recursion can be eliminated.

Proof: The book proves this for grammars that have no ε-productions and no cycles and has exercises asking the reader to prove that cycles and ε-productions can be eliminated.

We will try to avoid these hard cases.

Homework: Eliminate left recursion in the following grammar for simple postfix expressions.
X → S S + | S S * | a

4.3.4: Left Factoring

If two productions with the same LHS have their RHS beginning with the same symbol, then the FIRST sets will not be disjoint so predictive parsing (chapter 2) will be impossible and more generally top down parsing (later this chapter) will be more difficult as a longer lookahead will be needed to decide which production to use.

So convert A → x y1 | x y2 into

    A → x A'
   A' → y1 | y2
  
In other words factor out the x.

Homework: Left factor your answer to the previous homework.

4.3.5: Non-CFL Constructs

Although our grammars are powerful, they are not all-powerful. For example, we cannot write a grammar that checks that all variables are declared before used.

4.4: Top-Down Parsing

We did an example of top down parsing, namely predictive parsing, in chapter 2.

For top down parsing, we

  1. Start with the root of the parse tree, which is always the start symbol of the grammar. That is, initially the parse tree is just the start symbol.
  2. Choose a nonterminal in the frontier.
    1. Choose a production having that nonterminal as LHS.
    2. Expand the tree by making the RHS the children of the LHS.
  3. Repeat above until the frontier is all terminals.
  4. Hope that the frontier equals the input string.

The above has two nondeterministic choices (the nonterminal, and the production) and requires luck at the end. Indeed, the procedure will generate the entire language. So we have to be really lucky to get the input string.

4.4.1: Recursive Decent Parsing

Let's reduce the nondeterminism in the above algorithm by specifying which nonterminal to expand. Specifically, we do a depth-first (left to right) expansion.

We leave the choice of production nondeterministic.

We also process the terminals in the RHS, checking that they match the input. By doing the expansion depth-first, left to right, we ensure that we encounter the terminals in the order they will appear in the frontier of the final tree. Thus if the terminal does not match the corresponding input symbol now, it never will and the expansion so far will not produce the input string as desired.

Now our algorithm is

  1. Initially, the tree is the start symbol, the nonterminal we are processing.

  2. Choose a production having the current nonterminal A as LHS. Say the RHS is X1 X2 ... Xn.
  3. for i = 1 to n
      if Xi is a nonterminal
        process Xi  // recursive
      else if Xi (a terminal) matches current input symbol
        advance input to next symbol
      else // trouble Xi doesn't match and never will
          

Note that the trouble mentioned at the end of the algorithm does not signify an erroneous input. We may simply have chosen the wrong production in step 2.

In a general recursive descent (top-down) parser, we would support backtracking, that is when we hit the trouble, we would go back and choose another production. Since this is recursive, it is possible that no productions work for this nonterminal, because the wrong choice was made earlier.

The good news is that we will work with grammars where we can control the nondeterminism much better. Recall that for predictive parsing, the use of 1 symbol of lookahead made the algorithm fully deterministic, without backtracking.

4.4.2: FIRST and FOLLOW

We used FIRST(RHS) when we did predictive parsing.

Now we learn the whole truth about these two sets, which prove to be quite useful for several parsing techniques (and for error recovery).

The basic idea is that FIRST(α) tells you what the first symbol can be when you fully expand the string α and FOLLOW(A) tells what terminals can immediately follow the nonterminal A.

Definition: For any string α of grammar symbols, we define FIRST(α) to be the set of terminals that occur as the first symbol in a string derived from α. So, if α⇒*xQ for x a terminal and Q a string, then x is in FIRST(α). In addition if α⇒*ε, then ε is in FIRST(α).

Definition: For any nonterminal A, FOLLOW(A) is the set of terminals x, that can appear immediately to the right of A in a sentential form. Formally, it is the set of terminals x, such that S⇒*αAxβ. In addition, if A can be the rightmost symbol in a sentential form, the endmarker $ is in FOLLOW(A).

Note that there might have been symbols between A and x during the derivation, providing they all derived ε and eventually x immediately follows A.

Unfortunately, the algorithms for computing FIRST and FOLLOW are not as simple to state as the definition suggests, in large part caused by ε-productions.

  1. FIRST(a)={a} for all terminals a.
  2. Initialize FIRST(A)=φ for all nonterminals A
  3. If A → ε is a production, add ε to FIRST(A).
  4. For each production A → Y1 ... Yn, add to FIRST(A) any terminal a satisfying
    1. a is in FIRST(Yi) and
    2. ε is in all previous FIRST(Yj).
    Repeat this step until nothing is added.
  5. FIRST of any string X=X1X2...Xn is initialized to φ and then
    1. add to FIRST(X) any non-ε symbol in FIRST(Xi) if ε is in all previous FIRST(Xj).
    2. add ε to FIRST(X) if ε is in every FIRST(Xj).
  6. Initialize FOLLOW(S)=$ and FOLLOW(A)=φ for all other nonterminals A, and then apply the following three rules until nothing is add to any FOLLOW set.
    1. For every production A → α B β, add all of FIRST(β) except ε to FOLLOW(B).
    2. For every production A → α B, add all of FOLLOW(A) to FOLLOW(B).
    3. For every production A → α B β where FIRST(β) contains ε, add all of FOLLOW(A) to FOLLOW(B).

Do the FIRST and FOLLOW sets for

    E  → T E'
    E' → + T E' | ε
    T  → F T'
    T' → * F T' | ε
    F  → ( E ) | id
  

Homework: Compute FIRST and FOLLOW for the postfix grammar S → S S + | S S * | a

4.4.3: LL(1) Grammars

The predictive parsers of chapter 2 are recursive descent parsers needing no backtracking. A predictive parser can be constructed for any grammar in the class LL(1). The two Ls stand for (processing the input) Left to right and for producing Leftmost derivations. The 1 in parens indicates that 1 symbol of lookahead is used.

Definition: A grammar is LL(1) if for all production pairs A → α | β

  1. FIRST(α) ∩ FIRST(β) = φ.
  2. If β ⇒* ε, then no string derived from α begins with a terminal in FOLLOW(A). Similarly, if α ⇒* ε.

The 2nd condition may seem strange; it did to me for a while. Let's consider the simplest case that condition 2 is trying to avoid.

    S → A b    // b is in FOLLOW(A)
    A → b      // α=b so α derives a string beginning with b
    A → ε      // β=ε so β derives ε
  
ll1 def

Assume we are using predictive parsing and, as illustrated in the diagram to the right, we are at A in the parse tree and b in the input. Since lookahead=b and b is in FIRST(RHS) for the top A production, we would choose that production to expand A. But this could be wrong! Remember that we don't look ahead in the tree just in the input. So we would not have noticed that the next node in the tree (i.e., in the frontier) is b. This is possible since b is in FOLLOW(A). So perhaps we should use the second A production to produce ε in the tree, and then the next node b would match the input b.

Constructing a Predictive Parsing Table

The goal is to produce a table telling us at each situation which production to apply. A situation means a nonterminal in the parse tree and an input symbol in lookahead.

So we produce a table with rows corresponding to nonterminals and columns corresponding to input symbols (including $. the endmarker). In an entry we put the production to apply when we are in that situation.

We start with an empty table M and populate it as follows. (2nd edition has typo, A instead of α.) For each production A → α

  1. For each terminal a in FIRST(α), add A → α to M[A,a]. This is what we did with predictive parsing in chapter 2. The point was that if we are up to A in the tree and a is the lookahead, we should use the production A→α.
  2. If ε is in FIRST(α), then add A → α to M[A,b] (resp. M[A,$]) for each terminal b in FOLLOW(A) (if $ is in FOLLOW(A)). This is not so obvious; it corresponds to the second (strange) condition above. If ε is in FIRST(α), then α⇒*ε. Hence we should apply the production A→α, have the α go to ε and then the b (or $), which follows A will match the b in the input.

When we have finished filling in the table M, what do we do if an slot has

  1. no entries? This means that from this situation, no production is appropriate. Hence, if parsing an input leads to this entry, we cannot parse the sentence (because it is not in the language) so we report an error and try to repair it.
  2. one entry? Perfect! This means we know exactly what to in this situation.
  3. more than one entry? This should not happen since this section is entitled LL(1) grammars Someone erred when they said the grammar generated an LL(1) language. Since the language is not LL(1), we must use a different technique. One possibility is to use bottom-up parsing, which we study next. Another is to modify the procedure for this non-terminal to look further ahead (typically one more token) to decide what action to perform.

================ Start Lecture #6 ================

Example: Work out the parsing table for

    E  → T E'
    E' → + T E' | ε
    T  → F T'
    T' → * F T' | ε
    F  → ( E ) | id
  
FIRSTFOLLOW
E( id$ )
E'ε +$ )
T( id+ $ )
T'ε *+ $ )
F( id* + $ )

We already computed FIRST and FOLLOW as shown on the right. The table skeleton is

Nonter-
minal
Input Symbol
+*()id$
E
E'
T
T'
F

Homework: Produce the predictive parsing table for

  1. S → 0 S 1 | 0 1
  2. the prefix grammar S → + S S | * S S | a
Don't forget to eliminate left recursion and perform left factoring if necessary.

4.4.4: Nonrecursive Predictive Parsing

This illustrates the standard technique for eliminating recursion by keeping the stack explicitly. The runtime improvement can be considerable.

4.4.5: Error Recovery in Predictive Parsing

Skipped. bottom-up parse id*id

4.5: Bottom-Up Parsing

Now we start with the input string, i.e., the bottom (leaves) of what will become the parse tree, and work our way up to the start symbol.

For bottom up parsing, we are not as fearful of left recursion as we were with top down. Our first few examples will use the left recursive expression grammar

    E → E + T | T
    T → T * F | F
    F → ( E ) | id
  

4.5.1: Reductions

Remember that running a production in reverse, i.e., replacing the RHS by the LHS is called reducing. So our goal is to reduce the input string to the start symbol.

On the right is a movie of parsing id*id in a bottom-up fashion. Note the way it is written. For example, from step 1 to 2, we don't just put F above id*id. We draw it as we do because it is the current top of the tree (really forest) and not the bottom that we are working on so we want the top to be in horizontal line and hence easy to read.

The tops of the forest are the roots of the subtrees present in the diagram. For the movie those are
id * id, F * id, T * F, T, E
Note that (since the reduction successfully reaches the start symbol) each of these sets of roots is a sentential form.

The steps from one frame of the movie, when viewed going down the page, are reductions (replace the RHS of a production by the LHS). Naturally, when viewed going up the page, we have a derivation (replace LHS by RHS). For our example the derivation is
E ⇒ T ⇒ T * F ⇒ T * id ⇒ F * id ⇒ id * id

Note that this is a rightmost derivation and hence each of the sets of roots identified above is a right sentential form. So the reduction we did in the movie was a rightmost derivation in reverse.

Remember that for a non-ambiguous grammar there is only one rightmost derivation and hence there is only one rightmost derivation in reverse.

Remark: You cannot simply scan the string (the roots of the forest) from left to right and choose the first substring that matches the RHS of some production. If you try it in our movie you will reduce T to E right after T appears. The result is not a right sentential form.

Right
Sentential
Form
HandleReducing
Production
id1 * id2id1F → id
F * id2FT → F
T * id2id2F → id
T * FT * FE → T * F

4.5.2: Handle Pruning

The strings that are reduced during the reverse of a rightmost derivation are called the handles. For our example, this is shown in the table on the right.

Note that the string to the right of the handle must contain only terminals. If there was a non-terminal to the right, it would have been reduced in the RIGHTmost derivation that leads to this right sentential form.

Often instead of referring to a derivation A→α as a handle, we call α the handle. I should say a handle because there can be more than one if the grammar is ambiguous.

So (assuming a non-ambiguous grammar) the rightmost derivation in reverse can be obtained by constantly reducing the handle in the current string.

Homework: 4.23 a c

4.5.3: Shift-Reduce Parsing

We use two data structures for these parsers.

  1. A stack of grammar symbols, terminals and nonterminals. This stack is drawn in examples as having its top on the right and bottom on the left. The items shifted (see below) onto the stack will be terminals, but some are reduced to nonterminals. The bottom of the stack is marked with $ and initially the stack is empty (i.e., has just $).
  2. An input buffer that (conceptually) holds the remainder of the input, i.e., the part that has yet to be shifted onto the stack. An endmarker $ is placed after the end of the input. Initially the input buffer contains the entire input followed by $. (In practice we use some more sophisticated buffering technique, as we saw in section 3.2 with buffers pairs, that does not require having the entire input in memory at once.)

StackInputAction
$id1*id2$shift
$id1*id2$reduce F→id
$F*id2$reduce T→F
$T*id2$shift
$T*id2$shift
$T*id2$reduce F→id
$T*F$reduce T→T*F
$T$reduce E→T
$E$accept
The idea, illustrated by the table on the right, is that at any point the parser can perform one of four operations.
  1. The parser can shift a symbol from the beginning of the input onto the TOS.
  2. If the TOS is a handle, the parser can reduce it to its LHS.
  3. If the parser reaches the accepting state with the stack $S and the input $, the parser terminates successfully.
  4. The parser reaches an error state.

A technical point, which explains the usage of a stack is that a handle is always at the TOS. See the book for a proof; the idea is to look at what rightmost derivations can do (specifically two consecutive productions) and then trace back what the parser will do since it does the reverse operations (reductions) in the reverse order.

We have not yet discussed how to decide whether to shift or reduce when both are possible. We have also not discussed which reduction to choose if multiple reductions are possible. These are crucial question for bottom up (shift-reduce) parsing and will be addressed.

Homework: 4.23 b

4.5.4: Conflicts During Shift-Reduce Parsing

There are grammars (non-LR) for which no viable algorithm can decide whether to shift or reduce when both are possible or which reduction to perform when several are possible. However, for most languages, choosing a good lexer yields an LR(k) language of tokens. For example, ada uses () for both function calls and array references. If the lexer returned id for both array names and procedure names then a reduce/reduce conflict would occur when the stack was ... id ( id and the input ) ... since the id on TOS should be reduced to parameter if the first id was a procedure name and to expr if the first id was an array name. A better lexer (and an assumption, which is true in ada, that the declaration must precede the use) would return proc-id when it encounters a lexeme corresponding to a procedure name. It does this by constructing the symbol table it builds.

4.6: Introduction to LR Parsing: Simple LR

Remark: Both editions do a warm up before getting down to business with full LR parsing. The first edition does operator precedence and covers SLR in the middle of the section on LR parsing. The second omits operator precedence and does SLR here. I am following the second since operator precedence is no longer widely used and I believe SLR will be more helpful when trying to understand full LR and hence serves as a better introduction the subject.

Indeed, I will have much more to say about SLR than the other LR schemes. The reason is that SLR is simpler to understand, but does capture the essence of shift-reduce, bottom-up parsing. The disadvantage of SLR is that there are LR grammars that are not SLR.

I will just say the following about operator precedence. We shall see that a major consideration in all the bottom-up, shift-reduce parsers is deciding when to shift and when to reduct. Consider parsing A+B*C in C/java/etc. When the stack is A+B and the remaining input is *C, the parser needs to know whether to reduce A+B or shift in * and then C. (Really the A+B will probably by now be more like E+T.) The idea of operator precedence is that we give * higher precedence so when the parser see * on the input it knows not to reduce +. More details are in the first (i.e., your) edition of the text.

4.6.1: Why LR Parsers?

The text's presentation is somewhat controversial. Most commercial compilers use hand-written top-down parsers of the recursive-descent (LL not LR) variety. Since the grammars for these languages are not LL(1), the straightforward application of the techniques we have seen will not work. Instead the parsers actually look ahead further than one token, but only at those few places where the grammar is in fact not LL(1). Recall that (hand written) recursive descent compilers have a procedure for each nonterminal so we can customize as needed.

These compiler writers claim that they are able to produce much better error messages than can readily be obtained by going to LR (with its attendant requirement that a parser-generator be used since the parsers are too large to construct by hand). Note that compiler error messages is a very important user interface issue and that with recursive descent one can augment the procedure for a nonterminal with statements like
if (nextToken == X) then error(expected Y here)

Nonetheless, the claims made by the text are correct, namely.

  1. LR parsers can be constructed to recognize nearly all programming-language constructs for which CFGs exist.
  2. LR-parsing is the most general nonbacktracking, shift-reduce method known, yet can be implemented relatively efficiently.
  3. LR-parsing can detect a syntactic error as soon as possible.
  4. LR grammars can describe more languages than LL grammars.

4.6.2: Items and the LR(0) Automaton

We now come to grips with the big question: How does a shift-reduce parser know when to shift and when to reduce? This will take a while to answer in a satisfactory manner. The unsatisfactory answer is that the parser has tables that say in each situation whether to shift or reduce (or announce error, or announce acceptance). To begin the path toward the answer, we need several definitions.

An item is a production with a marker saying how far the parser has gotten with this production. Formally,

Definition: An (LR(0)) item of a grammar is a production with a dot added somewhere to the RHS.

Examples:

  1. E → E + T generates 4 items.
    1. E → · E + T
    2. E → E · + T
    3. E → E + · T
    4. E → E + T ·
  2. A → ε generates A → · as its only item.

The item E → E · + T signifies that the parser has just processed input that is derivable from E and will look for input derivable from + T.

Line 4 indicates that the parser has just seen the entire RHS and must consider reducing it to E. Important: consider does not mean do.

The parser groups certain items together into states. As we shall see, the items with a given state are treated similarly.

Our goal is to construct first the canonical LR(0) collection of states and then a DFA called the LR(0) automaton (technically not a DFA since no dead state).

To construct the canonical LR(0) collection formally and present the parsing algorithm in detail we shall

  1. augment the grammar
  2. define functions CLOSURE and GOTO

Augmenting the grammar is easy. We simply add a new start state S' and one production S'→S. The purpose is to detect success, which occurs when the parser is ready to reduce S to S'.

So our example grammar

    E → E + T | T
    T → T * F | F
    F → ( E ) | id
  
is augmented by adding the production E' → E.

Interlude: Thea Rough Idea

I hope the following interlude will prove helpful. In preparing to present SLR, I was struck how it looked like we were working with a DFA that came from some (unspecified and unmentioned) NFA. It seemed that by first doing the NFA, I could give some rough insight. Since for our current example the NFA has more states and hence a bigger diagram, let's consider the following extremely simple grammar.

    E → E + T
    E → T
    T → id
  
When augmented this becomes
   E' → E
    E → E + T
    E → T
    T → id
  
When the dots are added we get 10 items (4 from the second production, 2 each from the other three). See the diagram at the right. We begin at E'→.E since it is the start item. lr0 ajg nfa

Note that there are really four kinds of edges.

  1. Edges labeled with terminals. These correspond to shift actions, where the indicated terminal is shifted from the input to the stack.
  2. Edges labeled with nonterminals. These will correspond to reduce actions when we construct the DFA. The stack is reduced by a production having the given nonterminal as LHS. Reduce actions do more as we shall see.
  3. Edges labeled with ε. These are associated with the closure operation to be discussed and are the source of the nondeterminism (i.e., why the diagram is an NFA).
  4. An edge labeled $. This edge, which can be thought of as shifting the endmarker, is used when we are reducing via the E'→E production and accepting the input.

If we were at the item E→E·+T (the dot indicating that we have seen an E and now need a +) and shifted a + from the input to the stack we would move to the item E→E+·T. If the dot is before a non-terminal, the parser needs a reduction with that non-terminal as the LHS.

Now we come to the idea of closure, which I illustrate in the diagram with the ε's. Please note that this is rough, we are not doing regular expressions again, but I hope this will help you understand the idea of closure, which like ε in regular production leads to nondeterminism.

Look at the start state. The placement of the dot indicates that we next need to see an E. Since E is a nonterminal, we won't see it in the input, but will instead have to generate it via a production. Thus by looking for an E, we are also looking for any production that has E on the LHS. This is indicated by the two ε's leaving the top left box. Similarly, there are ε's leaving the other three boxes where the dot is immediately to the left of a nonterminal.

As with regular expressions, we combine n-items connected by an ε arc into a d-item. The actual terminology used is that we combine these items into a set of items (later referred to as a state). There is another combination that occurs. The top two n-items in the left column are combined into the same d-item and both n-items have E transitions (outgoing arcs labeled E). Since we are considering these two n-items to be the same d-item and the arcs correspond to the same transition, the two targets (the top two n-items in the 2nd column) are combined. A d-item has all the outgoing arcs of the original n-items it contains. This is the way we converted an NFAs into a DFA in the previous chapter.

I0, I1, etc are called (LR(0)) item sets, and the collection with the arcs (i.e., the DFA) is called the LR(0) automaton.

StackSymbolsInputAction
0id+id$Shift to 3
03id+id$Reduce by T→id
02T+id$Reduce by E→T.
01E+id$Shift to 4
014E+id$Shift to 3
0143E+id$Reduce by T→id
0145E+T$Reduce by E→E+T
01E$Accept
Now we put the diagram to use to parse id+id as shown in the table on the right. The symbols column is not needed since it can be determined from the stack, but it is useful for understanding. The first edition merges the stack and symbols columns, but I think it is clearer when they are separate as in the 2nd edition.

We start in the initial state with the stack empty and the input full. The $'s are just end markers. From state 0, called I0 in my diagram (following the book they are called I's since they are sets of items), we can only shift in the id (the nonterminals will appear in the symbols column). This brings us to I3 so we push a 3 onto the stack

In I3 we see a completed production in the box (the dot is on the extreme right). Thus we can reduce by this production. To reduce we pop the stack for each symbol in the RHS since we are replacing the RHS by the LHS; this time the RHS has one symbol so we pop the stack once and also remove one symbol. The stack corresponds to moves so we are undoing the move to 3 and we are temporarily in 0 again. But the production has a T on the LHS so we follow the T production from 0 to 2, push T onto Symbols, and push 2 onto the stack.

In I2 we again see a completed production and do another reduction, which brings us to 1.

The next two steps are shifts of + and id. We then reduce the id to T and are in step 5 ready for the big one.

The reduction in 5 has three symbols on the RHS so we pop (back up) three times again temporarily landing in 0, but the RHS puts us in 1.

Perfect! We have just E as a symbol and the input is empty so we are ready to reduce by E'→E, which signifies acceptance.

Now we rejoin the book and say it more formally.

================ Start Lecture #7 ================

Closure of Item Sets

Say I is a set of items and one of these items is A→α·Bβ. This item represents the parser having seen α and records that the parser might soon see the remainder of the RHS. For that to happen the parser must first see a string derivable from B. Now consider any production starting with B, say B→γ. If the parser is to making progress on A→α·Bβ, it will need to be making progress on one such B→·γ. Hence we want to add all the latter productions to any state that contains the former. We formalize this into the notion of closure.

Definition: For any set of items I, CLOSURE(I) is formed as follows.

  1. Initialize CLOSURE(I) = I
  2. If A → α · B β is in CLOSURE(I) and B → γ is a production, then add B → · γ to the closure and repeat.

Example: Recall our main example

   E' → E
    E → E + T | T
    T → T * F | F
    F → ( E ) | id
  
CLOSURE({E' → E}) contains 7 elements. The 6 new elements are the 6 original productions each with a dot right after the arrow.

The GOTO Function

If X is a grammar symbol, then moving from A→α·Xβ to A→αX·β signifies that the parser has just processed (input derivable from) X. The parser was in the former position and X was on the input; this caused the parser to go to the latter position. We (almost) indicate this by writing GOTO(A→α·Xβ,X) is A→αX·β. I said almost because GOTO is actually defined from item sets to item sets not from items to items.

Definition: If I is an item set and X is a grammar symbol, then GOTO(I,X) is the closure of the set of items A→αX·β where A→α·Xβ is in I.

The Canonical Collection of LR(0) Items

I really believe this is very clear, but I understand that the formalism makes it seem confusing. Let me begin with the idea.

We augment the grammar and get this one new production; take its closure. That is the first element of the collection; call it Z. Try GOTOing from Z, i.e., for each grammar symbol, consider GOTO(Z,X); each of these (almost) is another element of the collection. Now try GOTOing from each of these new elements of the collection, etc. Start with jane smith, add all her friends F, then add the friends of everyone in F, called FF, then add all the friends of everyone in FF, etc

The (almost) is because GOTO(Z,X) could be empty so formally we construct the canonical collection of LR(0) items, C, as follows

  1. Initialize C = CLOSURE({S' → S})
  2. If I is in C, X is a grammar symbol, and GOTO(I,X)≠φ then add it to C and repeat.

This GOTO gives exactly the arcs in the DFA I constructed earlier. The formal treatment does not include the NFA, but works with the DFA from the beginning.

Homework:

  1. Construct the LR(0) set of items for the following grammar (which produces simple postfix expressions).
    X → S S + | S S * | a
    Don't forget to augment the grammar.
  2. Draw the DFA for this item set.

lr0 4.31

The DFA for our Main Example

Our main example is larger than the toy I did before. The NFA would have 2+4+2+4+2+4+2=20 states (a production with k symbols on the RHS gives k+1 N-states since there k+1 places to place the dot). This gives rise to 11 D-states. However, the development in the book, which we are following now, constructs the DFA directly. The resulting diagram is on the right.

Start constructing the diagram on the board. Begin with {E' → ·E}, take the closure, and then keep applying GOTO.

4.6.3: The LR-Parsing Algorithm

The LR-parsing algorithm must decide when to shift and when to reduce (and in the latter case, by which production). It does this by consulting two tables, ACTION and GOTO. The basic algorithm is the same for all LR parsers, what changes are the tables ACTION and GOTO.

The LR Parsing Tables

We have already seen GOTO (for SLR).

Technical point that may, and probably should, be ignored: our GOTO was defined on pairs [item-set,grammar-symbol]. The new GOTO is defined on pairs [state,nonterminal]. A state (except the initial state) is an item set together with the grammar symbol that was used to generate it (via the old GOTO). We will not use the new GOTO on terminals so we just define it on nonterminals.

Given a state i and a terminal a (or the endmarker), ACTION[i,a] can be

  1. Shift j. The terminal a is shifted on to the stack and the parser enters state j.
  2. Reduce A → α. The parser reduces α on the TOS to A.
  3. Accept.
  4. Error

So ACTION is the key to deciding shift vs. reduce. We will soon see how this table is computed for SLR.

Since ACTION is defined on [state,terminal] pairs and GOTO is defined on [state,nonterminal], we can combine these tables into one defined on [state,grammar-symbol] pairs.

LR-Parser Configurations (formalism)

This formalism is useful for stating the actions of the parser precisely, but I believe it can be explained without it.

As mentioned above the Symbols column is redundant so a configuration of the parser consists of the current stack and the remainder of the input. Formally it is

(s0,s1...sm,aiai+1...an$
where the s's are states and the a's input symbols. This state could also be represented by the right-sentential form
X1...Xm,ai...an
where the X is the symbol associated with the state. All arcs into a state are labeled with this symbol. The initial state has no symbol.

Behavior of the LR Parser

The parser consults the combined ACTION-GOTO table for its current state (TOS) and next input symbol, formally this is ACTION[sm,ai], and proceeds as follows based on the value in the table. We have done this informally just above; here we use the formal treatment

  1. Shift s. The input symbol is pushed and becomes the new state. The new configuration is
    (s0...sms,ai+1...an
  2. Reduce A → α. Let r be the number of symbols in the RHS of the production. The parser pops r items off the stack (backing up r states) and enters the state GOTO(sm-r,A). That is after backing up it goes where A says to go. A real parser would now probably do something, e.g., a semantic action. Although we know about this from the chapter 2 overview, we don't officially know about it here. So for now simply print the production the parser reduced by.
  3. Accept.
  4. Error.

4.6.4: Constructing SLR-Parsing Tables

The missing piece of the puzzle is finally revealed.

A Terminology Point

The book (both editions) and the rest of the world seem to use GOTO for both the function defined on item sets and the derived function on states. As a result we will be defining GOTO in terms of GOTO. (I notice that the first edition uses goto for both; I have been following the second edition, which uses GOTO. I don't think this is a real problem.) Item sets are denoted by I or Ij, etc. States are denoted by s or si or (get ready) i. Indeed both books use i in this section. The advantage is that on the stack we placed integers (i.e., i's) so this is consistent. The disadvantage is that we are defining GOTO(i,A) in terms of GOTO(Ii,A), which looks confusing. Actually, we view the old GOTO as a function and the new one as an array (mathematically, they are the same) so we actually write GOTO(i,A) and GOTO[Ii,A].

The Table Construction Algorithm

We start with an augmented grammar (i.e., we added S' → S).

  1. Construct {I0,...,In} the LR(0) items.
  2. The parsing actions for state i.
    1. If A→α·bβ is in Ii for b a terminal, then ACTION[i,b]=shift j, where GOTO(Ii,b)=Ij.
    2. If A→α· is in Ii, for A≠S', then, for all b in FOLLOW(A), ACTION[i,b]=reduce A→α.
    3. If S'→S· is in Ii, then ACTION[I,$]=accept.
    4. If any conflicts occurred, the grammar is not SLR(1).
  3. If GOTO(Ii,A)=Ij, for a nonterminal A, then GOTO[i,A]=j.
  4. All entries not yet defined are error.
  5. The initial state is the one containing S'→·S.

StateACTIONGOTO
id+*()$ETF
0s5s4123
1s6acc
2r2s7r2r2
3r4r4r4r4
4s5s4823
5r6r6r6r6
6s5s493
7s5s410
8s6s11
9r1s7r1r1
10r3r3r3r3
11r5r5r5r5
Example: Our main example gives the table on the right. The entry s5 abbreviates shift and go to state 5.
The entry r2 abbreviates reduce by production number 2, where we have numbered the productions as follows.
  1. E → E + T
  2. E → T
  3. T → T * F
  4. T → F
  5. F → ( E )
  6. F → id

The shift actions can be read directly off the DFA. For example I1 with a + goes to I6, I6 with an id goes to I5, and I9 with a * goes to I7.

The reduce actions require FOLLOW. Consider I5={F→id·}. Since the dot is at the end, we are ready to reduce, but we must check if the next symbol can follow the F we are reducing to. Since FOLLOW(F)={+,*,),$}, in row 5 (for I5) we put r6 (for reduce by production 6) in the columns for +, *, ), and $.

The GOTO columns can also be read directly off the DFA. Since there is an E-transition (arc labeled E) from I0 to I1, the column labeled E in row 0 contains a 1.

Since the column labeled + is blank for row 7, we see that it would be an error if we arrived in state 7 when the next input character is +.

Finally, if we are in state 1 when the input is exhausted ($ is the next input character), then we have a successfully parsed the input.

StackSymbolsInputAction
0id*id+id$shift
05id*id+id$reduce by F→id
03F*id+id$reduct by T→id
02T*id+id$shift
027T*id+id$shift
0275T*id+id$reduce by F→id
027 10T*F+id$reduce by T→T*F
02T+id$reduce by E→T
01E+id$shift
016E+id$shift
0165E+id$reduce by F→id
0163E+F$reduce by T→F
0169E+T$reduce by E→E+T
01E$accept
Example: The diagram on the right shows the actions when SLR parsing id*id+id. On the blackboard let's do id+id*id and see how the precedence is handled.

Homework: Construct the SLR parsing table for the following grammar
X → S S + | S S * | a
You already constructed the LR(0) automaton for this example in the previous homework.

4.6.5: Viable Prefixes

Skipped.

4.7: More Powerful LR Parsers

We consider very briefly two alternatives to SLR, canonical-LR or LR, and lookahead-LR or LALR.

4.7.1: Canonical LR(1) Items

SLR used the LR(0) items, that is the items used were productions with an embedded dot, but contained no other (lookahead) information. The LR(1) items contain the same productions with embedded dots, but add a second component, which is a terminal (or $). This second component becomes important only when the dot is at the extreme right (indicating that a reduction can be made if the input symbol is in the appropriate FOLLOW set). For LR(1) we do that reduction only if the input symbol is exactly the second component of the item. This finer control of when to perform reductions, enables the parsing of a larger class of languages.

4.7.2: Constructing LR(1) Sets of Items

Skipped.

4.7.3: Canonical LR(1) Parsing Tables

Skipped.

4.7.4: Constructing LALR Parsing Tables

For LALR we merge various LR(1) item sets together, obtaining nearly the LR(0) item sets we used in SLR. LR(1) items have two components, the first, called the core, is a production with a dot; the second a terminal. For LALR we merge all the item sets that have the same cores by combining the 2nd components (thus permitting reductions when any of these terminals is the next input symbol). Thus we obtain the same number of states (item sets) as in SLR since only the cores distinguish item sets.

Unlike SLR, we limit reductions to occurring only for certain specified input symbols. LR(1) gives finer control; it is possible for the LALR merger to have reduce-reduce conflicts when the LR(1) items on which it is based is conflict free.

Although these conflicts are possible, they are rare and the size reduction from LR(1) to LALR is quite large. LALR is the current method of choice for bottom-up, shift-reduce parsing.

4.7.5: Efficient Construction of LALR Parsing Tables

Skipped.

4.7.6: Compaction of LR Parsing Tables

4.8: Using Ambiguous Grammars

Skipped.

4.8.1: Precedence and Associativity to Resolve Conflicts

Skipped.

4.8.2: The Dangling-Else Ambiguity

Skipped.

4.8.3: Error Recovery in LR Parsing

Skipped.

4.9: Parser Generators

4.9.1: The Parser Generator Yacc

The tool corresponding to Lex for parsing is yacc, which (at least originally) stood for yet another compiler compiler. This name is cute but somewhat misleading since yacc (like the previous compiler compilers) does not produce a compiler, just a parser.

The structure of the user input is similar to that for lex, but instead of regular definitions, one includes productions with semantic actions.

There are ways to specify associativity and precedence of operators. It is not done with multiple grammar symbols as in a pure parser, but more like declarations.

Use of Yacc requires a serious session with its manual.

4.9.2: Using Yacc with Ambiguous Grammars

Skipped.

Creating Yacc Lexical Analyzers with Lex

Skipped

4.9.4: Error Recovery in Yacc

Skipped

Chapter 5: Syntax-Directed Translation

Homework: Read Chapter 5.

Again we are redoing, more formally and completely, things we briefly discussed when breezing over chapter 2.

Recall that a syntax-directed definition (SDD) adds semantic rules to the productions of a grammar. For example to the production T → T1 / F we might add the rule
T.code = T1.code || F.code || '/'
if we were doing an infix to postfix translator.

Rather than constantly copying ever larger strings to finally output at the root of the tree after a depth first traversal, we can perform the output incrementally by embedding semantic actions within the productions themselves. The above example becomes
T → T1 / F { print '/' } Since we are generating postfix, the action comes at the end (after we have generated the subtrees for T1 and F, and hence performed their actions). In general the actions occur within the production, not necessarily after the last symbol.

For SDD's we conceptually need to have the entire tree available after the parse so that we can run the depth first traversal. (It is depth first since we are doing postfix; we will see other orders shortly.) Semantic actions can be performed during the parse, without saving the tree.

5.1: Syntax-Directed Definitions (SDDs)

Formally, attributes are values (of any type) that are associated with grammar symbols. Write X.a for the attribute a of symbol X. You can think of attributes as fields in a record/struct/object.

Semantic rules (rules for short) are associated with productions.

5.1.1: Inherited and Synthesized Attributes

Terminals can have synthesized attributes, that are given to it by the lexer (not the parser). There are no rules in an SDD giving values to attributes for terminals. Terminals do not have inherited attributes. A nonterminal A can have both inherited and synthesized attributes. The difference is how they are computed by rules associated with a production at a node N of the parse tree. We sometimes refer to the production at node N as production N.

Definition: A synthesized attribute of a nonterminal A, is defined at a node N, where A is the LHS. The attribute can depend only on (synthesized or inherited) attribute values at the children of N (the RHS of N) and on inherited attribute values at N itself.

The arithmetic division example above was synthesized.

ProductionSemantic Rules


L → E $L.val = E.val
E → E1 + TE.val = E1.val + T.val
E → E1 - TE.val = E1.val - T.val
E → TE.val = T.val
T → T1 * FT.val = T1.val * F.val
T → T1 / FT.val = T1.val / F.val
T → FT.val = F.val
F → ( E )F.val = E.val
F → numF.val = num.lexval

Example: The SDD at the right gives a left-recursive grammar for expressions with an extra nonterminal L added as the start symbol. The terminal num is given a value by the lexer, which corresponds to the value stored in the numbers table for lab 2.

Draw the parse tree for 7+6/3 on the board and verify that L.val is 9, the value of the expression.

Definition: This example use only synthesized attributes; such SDDs are called S-attributed and have the property that the rules give the attribute of the LHS in terms of attributes of the RHS.

Inherited attributes are more complicated since the node N of the parse tree with which it is associated (which is also the natural node to store the value) does not contain the production with the corresponding semantic rule.

Definition: An inherited attribute of a nonterminal B at node N (where B is the LHS) is defined by a semantic rule of the production at the parent of N (where B occurs in the RHS). The value depends only on attributes at N, N's siblings, and N's parent.

Note that when viewed from the parent node P (the site of the semantic rule), the inherited attribute depends on values at P and at P's children (the same as for synthesized attributes). However, and this is crucial, the nonterminal B is the LHS of a child of P and hence the attribute is naturally associated with that child. It is possibly stored there and is shown there in the diagrams below.

We will see an example with inherited attributes soon.

Definition:Often the attributes are just evaluations without side effects. In such cases we call the SDD an attribute grammar.

================ Start Lecture #8 ================

Remark: There was a question last time about SLR concerning B⇒*ε. Consider A→α·Bβ. Can we consider the dot to be on the other side of B since B derives ε? I said I thought not and want to add that, since B derives ε, these productions will appear in the LR(0) automaton and hence will be taken care of without any extra rules here.

Remark: Do 7+6/3 on board using the SDD from the end of the previous lecture (should have been done last time).

5.1.2: Evaluating an SDD at the Nodes of a Parse Tree

If we are given an SDD and a parse tree for a given sentence, we would like to evaluate the annotations at every node. Since, for synthesized annotations parents can depend on children, and for inherited annotations children can depend on parents, there is no guarantee that one can in fact find an order of evaluation. The simplest counterexample is the single production A→B with synthesized attribute A.syn, inherited attribute B.inh, and rules A.syn=B.inh and B.inh=A.syn+1. This means to evaluate A.syn at the parent node we need B.inh at the child and vice versa. Even worse it is very hard to tell, in general, if every sentence has a successful evaluation order.

All this not withstanding we will not have great difficulty because we will not be considering the general case.

Annotated Parse Trees

7-(6)

Recall that a parse tree has leaves that are terminals and internal nodes that are non-terminals. We when we decorate the parse tree with attributes, the result is called an annotated parse tree, which is constructed as follows. Each internal node corresponds to a production with the symbol labeling the node the LHS of the production. If there are no attributes for the LHS in this production, we leave the node as it was (I don't believe this is a common occurrence). If there are k attributes for the LHS, we replace the LHS in the parse tree by k equations. The LHS of the equation is the attribute and the right hand side is its value. Note that the annotated parse tree contains all the information of the original parse tree since we replaced something like E with something like E.att=7.

We computed the values to put in this tree for 7+6/3 and on the right is (7-6).

Homework: 5.1

Why Have Inherited Attributes?

3*5*4 left rec

Consider the following left-recursive grammar for multiplication of numbers and the parse tree on the right for 3*5*4.

  T → T * F
  T → F
  F → num

It is easy to see how the values can be propagated up the tree and the expression evaluated.

When doing top-down parsing, we need to avoid left recursion. Consider the grammar below, which is the result of removing the left recursion, and again its parse tree is shown on the right. Try not to look at the semantic rules for the moment. 3*5*4

ProductionSemantic RulesType



T → F T'T'.lval = F.valInherited
T.val = T'.tvalSynthesized



T' → * F T1' T'1.lval = T'.lval * F.valInherited
T'.tval = T'1.tvalSynthesized



T' → εT'.tval = T'.lvalSynthesized



F → numF.val = num.lexvalSynthesized

Now where on the tree should we do the multiplication 3*5? There is no node that has 3 and * and 5 as children. The second production is the one with the * so that is the natural candidate for the multiplication site. Make sure you see that this production (for 3*5) is associated with the blue highlighted node in the parse tree. The right operand (5) can be obtained from the F that is the middle child of this T'. F gets the value from its child, the number itself; this is an example of the simple synthesized case we have already seen, F.val=num.lexval (see the last semantic rule in the table).

But where is the left operand? It is located at the sibling of T' in the parse tree, i.e., at the F immediately to T's left. This F is not mentioned in the production associated with the T' node we are examining. So, how does T' get F.val from its sibling? The common parent, in this case T, can get the value from F and then our node can inherit the value from its parent.
Bingo! ... an inherited attribute. This can be accomplished by having the following two rules at the node T.
T.tmp = F.val
T'.lval = T.tmp

Since we have no other use for T.tmp, we combine the above two rules into the first rule in the table.

Now lets look at the second multiplication (3*5)*4, where the parent of T' is another T'. (This is the normal case. When there are n multiplies, n-1 have T' as parent and only one has T).

The red-highlighted T' is the site for the multiplication. However, it needs as left operand, the product 3*5 that its parent can calculate. So we have the parent (another T' node, the blue one in this case) calculate the product and store it as an attribute of its right child namely the red T'. That is the first rule for T' in the table.

We have now explained the first, third, and last semantic rules. These are enough to calculate the answer. Indeed, if we trace it through, 60 does get evaluated and stored in the bottom right T', the one associated with the ε-production. Our remaining goal is to get the value up to the root where it represents the evaluation of this term T and can be combined with other terms to get the value of a larger expression.

3*5*4 annotated

Going up is easy, just synthesize. I named the attribute tval, for term-value. It is generated at the ε-production from the lval attribute (which at this node is not a good name) and propagated back up. At the T node it is called simply val. At the right we see the annotated parse tree for this input.

Homework: Extend this SDD to handle the left-recursive, more complete expression evaluator given earlier in this section. Don't forget to eliminate the left recursion first.

It clearly requires some care to write the annotations.

Another question is how does the system figure out the evaluation order if one exists? That is the subject of the next section.

Remark: Consider the identifier table. The lexer creates it initially, but as the compiler performs semantic analysis and discover more information about various identifiers, e.g., type and visibility information, the table is updated. One could think of this is some inherited/synthesized attribute pair that during each phase of analysis is pushed down and back up the tree. However, it is not implemented this way; the table is made a global data structure that is simply updated. The the compiler writer must ensure manually that the updates are performed in an order respecting any dependences.

5.2: Evaluation Orders for SDD's

5.2.1: Dependency Graphs

dependency

The diagram on the right illustrates a great deal. The black shows the parse tree for the multiplication grammar just studied when applied to a single multiplication, e.g. 3*5. The synthesized attributes are shown in green and are written to the right of the grammar symbol at the node where they are defined. The inherited attributes are shown in red and are written to the left of the grammar symbol where it is defined.

Each green arrow points to the attribute calculated from the attribute at the tail of the arrow. These arrows either go up the tree one level or stay at a node. That is because a synthesized attribute can depend only on the node where it is defined and that node's children. The computation of the attribute is associated with the production at the node at its arrowhead. In this example, each synthesized attribute depends on only one other, but that is not required.

Each red arrow also points to the attribute calculated from the attribute at the tail. Note that two red arrows point to the same attribute. This indicates that the common attribute at the arrowheads, depends on both attributes at the tails. According to the rules for inherited attributes, these arrows either go down the tree one level, go from a node to a sibling, or stay within a node. The computation of the attribute is associated with the production at the parent of the node at the arrowhead.

5.2.2: Ordering the Evaluation of Attributes

The graph just drawn is called the dependency graph. In addition to being generally useful in recording the relations between attributes, it shows the evaluation order(s) that can be used. Since the attribute at the head of an arrow depends on the on the one at the tail, we must evaluate the head attribute after evaluating the tail attribute.

Thus what we need is to find an evaluation order respecting the arrows. This is called a topological sort. The rule is that the needed ordering can be found if and only if there are no (directed) cycles. The algorithm is simple.

  1. Choose a node having no incoming edges
  2. Delete the node and all incident edges.
  3. Repeat
If the algorithm terminates with nodes remaining, there is a directed cycle and no suitable evaluation order.

If the algorithm succeeds in deleting all the nodes, then the deletion order is a suitable evaluation order and there were no directed cycles.

Homework: The topological sort algorithm is nondeterministic (Choose a node) and hence there can be many topological sort orders. Find all the orders for the diagram above (you should label the nodes so you can describe the orders).

5.2.3: S-Attributed Definitions

Given an SDD and a parse tree, it is easy to tell (by doing a topological sort) whether a suitable evaluation exists (and to find one).

However, a very difficult problem is, given an SDD, are there any parse trees with cycles in their dependency graphs, i.e., are there suitable evaluation orders for all parse trees. Fortunately, there are classes of SDDs for which a suitable evaluation order is guaranteed.

As mentioned above an SDD is S-attributed if every attribute is synthesized. For these SDDs all attributes are calculated from attribute values at the children since the other possibility, the tail attribute is at the same node, is impossible since the tail attribute must be inherited for such arrows. Thus no cycles are possible and the attributes can be evaluated by a postorder traversal of the parse tree.

Since postorder corresponds to the actions of an LR parser when reducing the body of a production to its head, it is often convenient to evaluate synthesized attributes during an LR parse.

5.2.4 L-Attributed Definitions

Unfortunately, it is hard to live without inherited attributes. So we define a class that permits certain kinds of inherited attributes. l-attributed

Definition: An SDD is L-Attributed if each attribute is either

  1. Synthesized.
  2. Inherited from the left, and hence the name L-attributed.
    If the production is A → X1X2...Xn, then the inherited attributes for Xj can depend only on
    1. Inherited attributes of A, the LHS.
    2. Any attribute of X1, ..., Xj-1, i.e. only on symbols to the left of Xj.
  3. Attributes of Xj, *BUT* you must guarantee (separately) that these attributes do not by themselves cause a cycle.
l-attributed

Case three must be handled specially whenever it occurs. The top picture to the right illustrates what the first two cases look like and suggest why there cannot be any cycles. The picture below it corresponds to a fictitious R-attributed definition. One reason L-attributed definitions are favored over R, is the left to right ordering in English. See the example below on type declarations and also consider the grammars that result from left recursion.

Evaluating L-Attributed Definitions

The picture shows that there is an evaluation order for L-attributed definitions (again assuming no case 3). More formally, do a depth first traversal of the tree. The first time you visit a node, evaluate its inherited attributes (since you will know the value of everything it depends on), and the last time you visit it, evaluate the synthesized attributes. This is two-thirds of an Euler-tour traversal.

Homework: Suppose we have a production A → B C D. Each of the four nonterminals has two attributes s, which is synthesized, and i, which is inherited. For each set of rules below, tell whether the rules are consistent with (i) an S-attributed definition, (ii) an L-attributed definition, (iii) any evaluation order at all.

  1. A.s = B.i + C.i
  2. A.s = B.i + C.s and D.i = A.i + B.s
  3. A.s = B.s + D.s

5.2.5: Semantic Rules with Controlled Side Effects

ProductionSemantic RuleType



D → T LL.type = T.typeinherited
T → INTT.type = integersynthesized



L → L1 , ID L1.type = L.typeinherited
addType(ID.entry,L.type)synthesized, side effect



L → IDaddType(ID.entry,L.type)synthesized, side effect

When we have side effects such as printing or adding an entry to a table we must ensure that we have not added a constraint to the evaluation order that causes a cycle.

For example, the left-recursive SDD shown in the table on the right propagates type information from a declaration to entries in an identifier table.

The function addType adds the type information in the second argument to the identifier table entry specified in the first argument. Note that the side effect, adding the type info to the table, does not affect the evaluation order.

Draw the dependency graph on the board. Note that the terminal ID has an attribute (given by the lexer) entry that gives its entry in the identifier table. The nonterminal L has (in addition to L.type) a dummy synthesized attribute, say AddType, that is a place holder for the addType() routine. AddType depends on the arguments of addType(). Since the first argument is from a child, and the second is an inherited attribute of this node, we have legal dependences for a synthesized attribute.

Note that we have an L-attributed definition.

Homework: For the SDD above, give the annotated parse tree for

    INT a,b,c
  

================ Start Lecture #9 ================

Remark: See the new section Evaluating L-Attributed Definitions in section 5.2.4.

5.3: Applications of Syntax-Directed Translations

5.3.1: Construction of Syntax Trees

ProductionSemantic Rules


E → E 1 + T E.node = new Node('+',E1.node,T.node)
E → E 1 - T E.node = new Node('-',E1.node,T.node)
E → TE.node = T.node
T → ( E )T.node = E.node
T → IDT.node = new Leaf(ID,ID.entry)
T → NUMT.node = new Leaf(NUM,NUM.val)

Recall that in a syntax tree (technically an abstract syntax tree) we just have the essentials. For example 7+3*5, would have one + node, one *, and the three numbers. Lets see how to construct the syntax tree from an SDD.

Assume we have two functions Leaf(op,val) and Node(op,c1,...,cn), that create leaves and interior nodes respectively of the syntax tree. Leaf is called for terminals. Op is the label of the node (op for operation) and val is the lexical value of the token. Node is called for nonterminals and the ci's refer (are pointers) to the children.

ProductionSemantic RulesType



E → T E'E.node=E'.synSynthesized
E'node=T.nodeInherited



E' → + T E'1 E'1.node=new Node('+',E'.node,T.node)Inherited
E'.syn=E'1.synSynthesized



E' → - T E'1 E'1.node=new Node('-',E'.node,T.node)Inherited
E'.syn=E'1.synSynthesized



E' → εE'.syn=E'.nodeSynthesized
T → ( E )T.node=E.nodeSynthesized
T → IDT.node=new Leaf(ID,ID.entry)Synthesized
T → NUMT.node=new Leaf(NUM,NUM.val)Synthesized

The upper table on the right shows a left-recursive grammar that is S-attributed (so all attributes are synthesized).

Try this for x-2+y and see that we get the syntax tree.

When we eliminate the left recursion, we get the lower table on the right. It is a good illustration of dependencies. Follow it through and see that you get the same syntax tree as for the left-recursive version.

Remarks:

  1. In the first edition (section 8.1) we have nearly the same table. The main difference is the switch from algol/pascal-like notation (mknode) to a java/object-like new.
  2. These two functions, new Node and new Leaf (or their equivalent), are needed for lab 3 (part 4), if you are doing a recursive-descent parser. When processing a production
    1. Create a parse tree node for the LHS.
    2. Call subroutines for RHS symbols and connect the resulting nodes to the node created in i.
    3. Return a reference to the new node so the parent can hook it into the parse tree.
  3. It is the lack of a call to new in the third and fourth productions that causes the (abstract) syntax tree to be produced rather than the parse (concrete syntax) tree.
  4. Production compilers do not produce a parse tree, but only the syntax tree. The syntax tree is smaller, and hence more (space and time) efficient for subsequent passes that walk the tree. The parse tree might be (I believe) very slightly easier to construct as you don't have to decide which nodes to produce; you simply produce them all.

5.3.2: The structure of a Type

This course emphasizes top-down parsing (at least for the labs) and hence we must eliminate left recursion. The resulting grammars need inherited attributes, since operations and operands are in different productions. But sometimes the language itself demands inherited attributes. Consider two ways to describe a 3x4, two-dimensional array. tree rep for arrays

    array [3] of array [4] of int    and     int[3][4]
  

Assume that we want to produce a structure like the one the right for the array declaration given above. This structure is generated by calling a function array(num,type). Our job is to create an SDD so that the function gets called with the correct arguments.

For the first language representation of arrays (found in Ada and similar to that in lab 3), it is easy to generate an S-attributed (non-left-recursive) grammar based on
A → ARRAY [ NUM ] OF A | INT | FLOAT
This is shown in the table on the left.

ProductionSemantic RulesType



T → B CT.t=C.tSynthesized
C.b=B.tInherited



B → INTB.t=integerSynthesized
B → FLOATB.t=floatSynthesized



C → [ NUM ] C1 C.t=array(NUM.val,C1.t)Synthesized
C1.b=C.bInherited



C → εC.t=C.bSynthesized
ProductionSemantic Rule


A → ARRAY [ NUM ] OF A1 A.t=array(NUM.val,A1.t)
A → INTA.t=integer
A → FLOATA.t=float

On the board draw the parse tree and see that simple synthesized attributes above suffice.

For the second language representation of arrays (the C-style), we need some smarts (and some inherited attributes) to move the int all the way to the right. Fortunately, the result, shown in the table on the right, is L-attributed and therefore all is well.

Homework: 5.6

5.4: Syntax-Directed Translation Schemes (SDTs)

Basically skipped.

The idea is that instead of the SDD approach, which requires that we build a parse tree and then perform the semantic rules in an order determined by the dependency graph, we can attach semantic actions to the grammar (as in chapter 2) and perform these actions during parsing, thus saving the construction of the parse tree.

But except for very simple languages, the tree cannot be eliminated. Modern commercial quality compilers all make multiple passes over the tree, which is actually the syntax tree (technically, the abstract syntax tree) rather than the parse tree (the concrete syntax tree).

5.4.1: Postfix Translation Schemes

If parsing is done bottom up and the SDD is S-attributed, one can generate an SDT with the actions at the end (hence, postfix). In this case the action is perform at the same time as the RHS is reduced to the LHS.

5.4.2: Parser-Stack Implementation of Postfix SDTs

Skipped.

5.4.3: SDTs with Actions Inside Productions

Skipped

5.4.4: Eliminating Left Recursion from SDTs

Skipped

5.4.5: SDTs For L-Attributed Definitions

Skipped

5.5: Implementing L-Attributed SDD's

A good summary of the available techniques.

  1. Build the parse tree and annotate. Works as long as no cycles are present (guaranteed by L- or S-attributed).
  2. Build the parse tree, add actions, and execute the actions in preorder. Works for any L-attributed definition. Can add actions based on the semantic rules of the SDD.
  3. Translate During Recursive Descent Parsing. See below.
  4. Generate Code on the Fly. Also uses recursive descent, but is restrictive.
  5. Implement an SDT during LL-parsing. Skipped.
  6. Implement an SDT during LR-parsing of an LL Language. Skipped.

5.5.1: Translation During Recursive-Descent Parsing

Recall that in recursive-descent parsing there is one procedure for each nonterminal. Assume the SDD is L-attributed. Pass the procedure the inherited attributes it might need (different productions with the same LHS need different attributes). The procedure keeps variables for attributes that will be needed (inherited for nonterminals in the body; synthesized for the head). Call the procedures for the nonterminals. Return all synthesized attributes for this nonterminal.

5.5.2: On-the-fly Code Generation

5.5.3: L-attributed SDDs and LL Parsing

5.5.4: Bottom-Up Parsing of L-Attributed SDDs

Requires an LL (not just LR) language.

What is this all used for?

Assume we have a parse tree as produced, for example, by your lab3. You now want to write the semantics analyzer, or intermediate code generator, and you have these semantic rules or actions that need to be performed. Assume the grammar is L-attributed, so we don't have to worry about dependence loops.

You start to write

analyze (tree-node)
This procedure is basically a big switch statement where the cases correspond to the different productions in the grammar. The tree-node is the LHS of the production and the children are the RHS. So by first switching on the tree-node and then inspecting enough of the children, you can tell the production.

As described in 5.5.1 above, you have received as parameters (in addition to tree-node), the attributes you are to inherit. You then call yourself recursively, with the tree-node argument set to your leftmost child, then call again using the next child, etc. Each time, you pass to the child the attributes it needs to inherit (You may be giving it too many since you know the nonterminal represented by this child but not the production; you could find out the production by examining the child's children, but probably don't bother doing so.)

When each child returns, it supplies as its return value the synthesized attributes it is passing back to you.

After the last child returns, you return to your caller, passing back the synthesized attributes you are to calculate.

Variations

  1. Instead of a giant switch, you could have separate routines for each nonterminal as done in the parser and just switch on the productions having this nonterminal as LHS.
  2. You could have separate routines for each production (requires looking 2-deep, as mentioned above).
  3. If you like actions instead of rules, perform the actions where indicated in the SDT.
  4. Global variable can be used (with care) instead of parameters.
  5. As illustrated earlier in the notes, you can call routines instead of setting an attribute (see addType in 5.2.5).

Chapter 6: Intermediate-Code Generation

Remark: This corresponds to chapters 6 and 8 in the first edition. The change is that storage management is now done after intermediate code generation.

Homework: Read Chapters 6 and 8.

6.1: Variants of Syntax Trees

Remark: This is 8.1 in 1e.

6.1.1: Directed Acyclic Graphs for Expressions

The difference between a syntax DAG and a syntax tree is that the former can have undirected cycles. DAGs are useful where there are multiple, identical portions in a given input. The common case of this is for expressions where there often are common subexpressions. For example in the expression
X + a + b + c - X + ( a + b + c )
each individual variable is a common subexpression. But a+b+c is not since the first occurrence has the X already added. This is a real difference when one considers the possibility of overflow or of loss of precision. The easy case is
x + y * z * w - ( q + y * z * w )
where y*z*w is a common subexpression.

It is easy to find these. The constructor Node() above checks if an identical node exists before creating a new one. So Node ('/',left,right) first checks if there is a node with op='/' and children left and right. If so, a reference to that node is returned; if not, a new node is created as before.

Homework: Construct the DAG for
((x+y)-((x+y)*(x-y)))+((x+y)*(x-y))

6.1.2: The Value-Number Method for Constructing DAGS

Often one stores the tree or DAG in an array, one entry per node. Then references to the array index of a node is called the node's value-number. Searching an unordered array is slow; there are many better data structures to use. Hash tables are a good choice.

6.2: Three-Address Code

Instructions of the form op a,b,c, where op is a primitive operator. For example

    lshift a,b,4   // left shift b by 4 and place result in a
    add    a,b,c   // a = b + c
    a = b + c      // alternate (more natural) representation of above
  

If we are starting with a DAG (or syntax tree if less aggressive), then transforming into 3-address code is just a topological sort and an assignment of a 3-address operation with a new name for the result to each interior node (the leaves already have names and values).

3 address code

For example, (B+A)*(Y-(B+A)) produces the DAG on the right, which yields the following 3-address code.

    t1 = B + A
    t2 = Y - t1
    t3 = t1 * t2
  

6.2.1: Addresses and Instructions

We use the term 3-address when we view the (intermediate-) code as having one elementary operation with three operands, each of which is an address. Typically two of the addresses represent source operands or arguments of the operation and the third represents the result. Some of the 3-address operations have fewer than three addresses; we simply think of the missing addresses as unused (or ignored) fields in the instruction.

Possible addresses

  1. (Source program) Names. Really the intermediate code would contain a reference to the (identifier) table entry for the name. For convenience, the actually identifier is often written.
  2. Constants. Again, this would often be a reference to a table entry. An important issue is type conversion that will be discussed later. Type conversion also applies to identifiers.
  3. (Compiler-generated) Temporaries. Although it may at first seem wasteful, modern practice assigns a new name to each temporary, rather than reusing the same temporary. (Remember that a DAG node is considered one temporary even if it has many parents.) Later phases can combine several temporaries into one (e.g., if they have disjoint lifetimes).

Possible three-address instructions

There is no universally agreed to set of three-address instructions or to whether 3-address code should be the intermediate code for the compiler. Some prefer a set close to a machine architecture. Others prefer a higher-level set closer to the source, for example, subsets of C have been used. Others prefer to have multiple levels of intermediate code in the compiler with one phase of compilation being converting from the high-level intermediate code into the low-level intermediate code. What follows is the set proposed in the 2ed; it looks to be essentially the same as that in the 1e.

In the list below, x, y, and z are addresses, i is an integer, and L is a symbolic label, as used in chapter 2. The instructions can be thought of as numbered and the labels can be converted to the numbers with another pass over the output or via backpatching, which is discussed below.

  1. Binary ops. x = y op z
  2. Unary ops. x = op y (includes copy, where op is the identity f(x)=x)
  3. Junp. goto L.
  4. Conditional unary op jumps. if x goto L   ifFalse x goto L.
  5. Conditional binary op jumps. if x relop y goto L
  6. Procedure/Function Calls and Returns.
      param x     call p,n     y = call p,n     return     return y.
  7. Indexed Copy ops. x = y[i]   x[i] = y.
  8. Address and pointer ops. x = &y   x = *y   *x

Homework: 8.1

6.2.2: Quadruples (Quads)

An easy way to represent the three address instructions: put the op into the first of four fields and the addresses into the remaining three. Some instructions do not use all the fields. Many operands will be references to entries in tables (e.g., the identifier table).

6.2.3: (Indirect) Triples

Optimization to save a field. The result field of a quad is omitted in a triple since the result is often a temporary.

When this result occurs as a source operand of a subsequent instruction, we indicate it by writing the value-number of the first instruction (distinguished some way, say with parens) as the operand of the second.

If the result field of a quad is not a temporary then two triples may be needed: One to do the operation and place the result into a temporary (which is not a field of the instruction). The second operation is a copy operation from the temporary to the final home. Recall that a copy does not use all the fields of a quad no fits into a triple without omitting the result.

When an optimizing compiler reorders instructions for increased performance, extra work is needed with triples since the instruction numbers, which have changed, are used implicitly. Hence the triples must be regenerated with correct numbers as operands.

Indirect triples. Keep an array of pointers to triples and, if it is necessary to reorder instructions, just reorder these pointers. This has two advantages.

  1. The pointers are (probably) smaller than the triples so faster to move. This is a generic advantage and could be used for quads and many other reordering applications (e.g., sorting large records).
  2. Since the triples don't move, the references they contain to past results remain accurate. This is specific to triples (or similar situations).

Homework: 8.2

6.2.4: Static Single-Assignment (SSA) Form

This has become a big deal in modern optimizers, but we will largely ignore it. The idea is that you have all assignments go to unique (temporary) variables. So if the code is
if x then y=4 else y=5
it is treated as though it was
if x then y1=4 else y2=5
The interesting part comes when y is used later in the program and the compiler must choose between y1 and y2.

6.3: Types and Declarations

Much of the early part of this section is really programming languages. In 1e this is section 6.1 (back from chapter 8).

6.3.1: Type Expressions

A type expression is either a basic type or the result of applying a type constructor.

Definition: A type expression is one of the following.

  1. A basic type.
  2. A type name.
  3. Applying an array constructor array(number,type-expression). In 1e, the number argument is an index set. This is where the C/java syntax is, in my view, inferior to the more algol-like syntax of e.g., ada and lab 3
    array [ index-type ] of type.
  4. Applying a record constructor record(field names and types).
  5. Applying a function constructor type→type.
  6. The product type×type.
  7. A type expression may contain variables (that are type expressions).

6.3.2: Type Equivalence

There are two camps, name equivalence and structural equivalence.

Consider the following for example.

    declare
       type MyInteger is new Integer;
       MyX : MyInteger;
       x   : Integer := 0;
    begin
       MyX := x;
    end
  
This generates a type error in Ada, which has name equivalence since the types of x and MyX do not have the same name, although they have the same structure.

When you have an object of an anonymous type as in
    x : array [5] of integer;
it doesn't have the same type as any other object even
    y : array [5] of integer;
But x[2] has the same type as y[3]; both are integers.

6.3.3: Declarations

The following from 2ed uses C/Java array notation. The 1ed has pascal-like material (section 6.2). Although I prefer Ada-like constructs as in lab 3, I realize that the class knows C/Java best so like the authors I will go with the 2ed. I will try to give lab3-like grammars as well.

This grammar gives C/Java like records/structs/methodless-classes as well as multidimensional arrays (really arrays of arrays).

    D → T id ; D | ε
    T → B C | RECORD { D }
    B → INT | FLOAT
    C → [ NUM ] C | ε
  

The lab 3 grammar doesn't support records and the support for multidimensional arrays is flawed (you can define the type, but not a (constrained) object). Here is the part of the lab3 grammar that handles declarations of ints, reals and arrays.

    declarations         → declaration declarations | ε
    declaration          → object-declaration | type-declaration
    object-declaration   → defining-identifier : object-definition ;
    object-definition    → type-name | type-name [ NUMBER ]
    type-declaration     → TYPE defining-identifier IS ARRAY OF type-name ;
    defining-identifier  → IDENTIFIER
    type-name            → IDENTIFIER | INT | REAL
  
So that the tables below are not too wide, let's use shorter names
    ds   → d ds | ε
    d    → od | td
    od   → di : odef ;
    odef → tn | tn [ NUM ]
    td   → TYPE di IS ARRAY OF tn ;
    di   → ID
    tn   → ID | INT | REAL
  

Ada Constrained vs Unconstrained Array Types (unofficial)

Ada supports both constrained array types such as
    type t1 is array [5] of integer
and unconstrained array types (as in lab 3) such as
    type t2 is array of integer
With the latter, the constraint is specified when the array (object) itself is declared.
    x1 : t1
    x2 : t2[5]
The grammar in lab3 supports t2 and x2, but not t1 and x1. The deficiency of the lab3 grammar is that for two dimensional array types
    type t3 is array of t2
we have no way to supply the two array bounds in the array (object) definition. Ada, which as said above, has both constrained and unconstrained array types, forbids the latter from appearing after is array of.

You might wonder why we want the unconstrained type. These types permit a procedure to have a parameter that is an array of integers of unspecified size. Remember that the declaration of a procedure specifies only the type of the parameter; the object is determined at the time of the procedure call.

6.3.4: Storage Layout for Local Names

See section 8.2 in 1e (we are going back to chapter 8 from 6, so perhaps Doc Brown from BTTF should give the lecture).

We are considering here only those types for which the storage can be computed at compile time. For others, e.g., string variables, dynamic arrays, etc, we would only be reserving space for a pointer to the structure; the structure itself is created at run time and is discussed in the next chapter.

The idea is that the basic type determines the width of the data, and the size of an array determines the height. These are then multiplied to get the size (area) of the data.

The book uses semantic actions (i.e., a syntax directed translation SDT). I added the corresponding semantic rules so that we have an SDD as well.

Remember that for an SDT, the placement of the actions withing the production is important. Since it aids reading to have the actions lined up in a column, we sometimes write the production itself on multiple lines. For example the production T→BC has the B and C on separate lines so that the action can be in between even though it is written to the right of both.

The actions use global variables t and w to carry the base type (INT or FLOAT) and width down to the ε-production, where they are then sent on their way up and become multiplied by the various dimensions. In the rules I use inherited attributes bt and bw. This is similar to the comment above that instead of having the identifier table passed up and down via attributes, the bullet is bitten and a globally visible table is used.

The base types and widths are set by the lexer or are constants in the parser.

ProductionActionsSemantic RulesKind




T → B { t = B.type; w = B.width; } C.bt = B.btInherited
        C { T.type = C.type; T.width = B.width; }




B → INT{ B.type = integer; B.width = 4; } B.bt = integer
B.bw = 4
Synthesized
Synthesized




B → FLOAT{ B.type = float; B.width = 8; } B.bt = integer
B.bw = 8
Synthesized
Synthesized




C → [ NUM ] C1 C.type = array(NUM.value, C1.type) Synthesized
C.width = NUM.value * C1.width; Synthesized
{ C.type = array(NUM.value, C1.type); C1.bt = C.bt Inherited
C.width = NUM.value * C1.width; } C1.bw = C.bw Inherited




C → εC.type = t; C.width=w C.type = C.bt
C.width = C.bw
Synthesized
Synthesized

Using the Lab 3 Grammar

Scalar Declarations
ProductionSemantic Rules


d → odd.width = od.width
d → tdd.width = 0


od → di : odef ;addType(di.entry, odef.type)
od.width = odef.width


di → IDdi.entry = ID.entry


odef → tnodef.type = tn.type
odef.width = tn.width
tn.type must be integer or real


tn → INTtn.type = integer
tn.width = 4


tn → REALtn.type = real
tn.width = 8

First let's ignore arrays. Then we get the simple table on the right. All the attributes are Synthesized so we have an S-attributed grammar.

We dutifully synthesize the width attribute all the way to the top and then do not use it. We shall use it in the next section when we consider multiple declarations.

Recall that addType is viewed as a synthesized since its parameters come from the RHS, i.e., from children of this node. It has a side effect (of modifying the identifier table) so we must be sure that we are not depending on some order of evaluation that is not simply parent after children. In fact, later when we evaluate expressions, we will need some of this information. We will need to enforce declaration before use since we will be looking up information that we are setting here. So in evaluation, we check the entry in the identifier table to be sure that the type (for example) has already been set.

Note the comment tn.type must be integer or real. This is an example of a type check, a key component of semantic analysis, that we will learn about soon. The reason for it here is that we are only able to handle 1 dimensional arrays with the lab3 grammar. (It would be a more complicated grammar with other type check rules to handle the general case found in ada).

================ Start Lecture #10 ================
scalar declaration

On the board, construct the parse tree, starting from the declaration for
      y: int ;
We should get the diagram on the right.

Now let's consider arrays. We need to include td (type-definition) and tn (type-name) as well as an additional production for od (object-definition). For td we need the restriction that tn is a basic type since we cannot define higher dimensional arrays.

I put in many type checks to distinguish the array case from the scalar case; possibly some are superfluous.

Once again all attributes are synthesized (including those with side effects) so we have an S-attributed SDD.

Array Declarations
ProductionSemantic Rules


d → odd.width = od.width
d → tdd.width = 0


od → di : odef ; addType(di.entry, odef.type)
od.width = odef.width


di → IDdi.entry = ID.entry


odef → tnodef.type = tn.type
odef.width = tn.width
tn.type must be integer or real


odef → tn [ NUM ] odef.type = array(NUM.value, getBaseType(tn.entry.type)
odef.width = sizeof(odef.type)
    = NUM.value*sizeof(getBaseType(tn.entry.type))
tn must be ID


td → TYPE di IS ARRAY OF tn ; addType(di.entry, Array(*, tn.type))
tn.type must be integer or real


tn → IDtn.entry = ID.entry
ID.entry.type must be array()


tn → INTtn.type = integer
tn.width = 4


tn → REALtn.type = real
tn.width = 8
array declaration

The top diagram on the right shows the result of applying the semantic actions in the table above to to the declaration
      type t is array of real;

The middle diagram shows the result after
      x : t[10];

The diagram below parses the following program using the lab grammar. Actually the diagram cheats since the lab grammar requires a statement and the diagram pretends that statement can be ε.

  Procedure P1 is
    y : integer;
    type t is array of real ;
    x : t[10];
declarations in a procedure


6.3.5: Sequences of Declarations

The Run Time Storage of Objects

Be careful to distinguish three methods used to store and pass information.

  1. Atrributes. These are variables in a phase (semantics analyzer; also intermediate code generator) of the compiler.
  2. Identifier (and other) table. This is longer lived data; often passed between phases.
  3. Run time storage. This is storage established by the compiler, but not used by the compiler. It is allocated and used during run time.

To summarize, the identifier table (and others we have used) are not present when the program is run. But there must be run time storage for objects. We need to know the address each object will have during execution. Specifically, we need to know its offset from the start of the area used for object storage.

For just one object, it is trivial: the offset is zero.

Multiple Declarations

The goal is to permit multiple declarations in the same procedure (or program or function). For C/java like languages this can occur in two ways.

  1. Multiple objects in a single declaration.
  2. Multiple declarations in a single procedure.

In either case we need to associate with the object being declared its storage location. Specifically we include in the table entry for the object, its offset from the beginning of the current procedure. We initialize this offset at the beginning of the procedure and increment it after each object declaration.

The programming languages Ada and Pascal do not permit multiple objects in a single declaration. Both languages are of the
    object : type
school. Thus lab 3, which follows Ada, and 1e, which follows pascal, do not support multiple objects in a single declaration. C/Java certainly does permit multiple objects, but surprisingly the 2e grammar does not.

Naturally, the way to permit multiple declarations is to have a list of declarations in the natural right-recursive way. The 2e C/Java grammar has D which is a list of semicolon-separated T ID's
    D → T ID ; D | ε

The lab 3 grammar has a list of declarations (each of which ends in a semicolon). Shortening declarations to ds we have
    ds → d ds | ε

As mentioned, we need to maintain an offset, the next storage location to be used by an object declaration. The 2e snippet below introduces a nonterminal P for program that gives a convenient place to initialize offset.

    P →                { offset = 0; }
        D
    D → T ID ;         { top.put(id.lexeme, T.type, offset);
                              offset = offset + T.width; }
        D1
    D → ε
  

The name top is used to signify that we work with the top symbol table (when we have nested scopes for record definitions we need a stack of symbol tables). Top.put places the identifier into this table with its type and storage location. We then bump offset for the next variable or next declaration.

Rather that figure out how to put this snippet together with the previous 2e code that handled arrays, we will just present the snippets and put everything together on the lab 3 grammar.

In the function-def (fd) and procedure-def (pd) productions we add the inherited attribute offset to declarations (ds.offset) and set it to zero. We then inherit this offset down to an individual declaration. If this is an object declaration, we store it in the entry for the identifier being declared and we increment the offset by the size of this object. When we get the to the end of the declarations (the ε-production), the offset value is the total size needed. So we turn it around and send it back up the tree.

Multiple Declarations
ProductionSemantic RulesKind



fd → FUNC di ( ps ) RET tn IS ds BEG s ss END ; ds.offset = 0 Inherited



pd → PROC di ( ps ) IS ds BEG s ss END ; ds.offset = 0 Inherited
s.next = newlabel() Inherited
ss.next = newlabel() Inherited
pd.code = s.code || label(s.next) || ss.code || label(ss.next) Synthesized



ds → d ds1 d.offset = ds.offset Inherited
ds1.offset = d.newoffset Inherited
ds.totalSize = ds1.totalSize Synthesized



ds → ε ds.totalSize = ds.offset Synthesized



d → od od.offset = d.offset Inherited
d.newoffset = d.offset + od.width Synthesized



d → td d.newoffset = d.offset Synthesized



od → di : odef ; addType(di.entry, odef.type) Synthesized
od.width = odef.width Synthesized
addOffset(di.entry, od.offset) Synthesized



di → ID di.entry = ID.entry Synthesized



odef → tn odef.type = tn.type Synthesized
odef.width = tn.width Synthesized
tn.type must be integer or real



odef → tn [ NUM ] odef.type = array(NUM.value, getBaseType(tn.entry.type)) Synthesized
odef.width = sizeof(odef.type) Synthesized
tn must be ID



td → TYPE di is ARRAY OF tn ; addType(di.entry, array(*, tn.type)) Synthesized
tn.type must be integer or real



tn → ID tn.entry = ID.entry Synthesized
ID.entry.type must be array()



tn → INT tn.type = integer Synthesized
tn.width = 4 Synthesized



tn → REAL tn.type = real Synthesized
tn.width = 8 Synthesized

Now show what happens when the following program is parsed and the semantic rules above are applied.

    procedure test () is
        y : integer;
        type t is array of real;
        x : t[10];
    begin
        y = 5;        // we haven't yet done statements
        x[2] = y;     // type error?
    end;
  

6.3.6: Fields in Records and Classes

Since records can essentially have a bunch of declarations inside, we only need add
T → RECORD { D }
to get the syntax right. For the semantics we need to push the environment and offset onto stacks since the namespace inside a record is distinct from that on the outside. The width of the record itself is the final value of (the inner) offset.

    T → record {         { Env.push(top);  top = new Env()
                           Stack.puch(offset); offset = 0; }
    D }                  { T.type = record(top); T.width = offset;
                           top = Env.pop(); offset = Stack.pop(); }
  

This does not apply directly to the lab 3 grammar since the grammar does not have records. It does, however, have procedures that can be nested. If we wanted to generate code for nested procedures we would need to stack the symbol table as done here in 2e.

Homework: Determine the types and relative addresses for the identifiers in the following sequence of declarations.

   float x;
   record { float x; float y; } rec;
   float y;
 

6.4: Translation of Expressions

Remark: See 8.3 in 1e.

Assignment Statements Without Arrays
ProductionSemantic Rule


as → lv = e as.code = e.code || gen(lv.lexeme = e.addr)


lv → ID lv.lexeme = get(ID.lexeme)


e → t e.addr = t.addr
e.code = t.code


e → e1 + t e.addr = new Temp()
e.code = e1.code || t.code || gen(e.addr = e1.addr + t.addr)


e → e1 - t e.addr = new Temp()
e.code = e1.code || t.code || gen(e.addr = e1.addr - t.addr)


t → f t.addr = f.addr
t.code = f.code


t → t1 * f t.addr = new Temp()
t.code = t1.code || f.code || gen(t.addr = t1.addr * f.addr)


t → t1 / f t.addr = new Temp()
t.code = t1.code || f.code || gen(t.addr = t1.addr / f.addr)


f → ( e ) f.addr = e.addr
f.code = e.code


f → ID f.addr = get(ID.lexeme)
f.code = ""


f → NUM f.addr = get(NUM.lexeme)
f.code = ""

The goal is to generate 3-address code for expressions. We will generate them using the natural notation of 6.2. In fact we assume there is a function gen() that given the pieces needed does the proper formatting so gen(x = y + z) will output the corresponding 3-address code. gen() is often called with addresses rather than lexemes like x. The constructor Temp() produces a new address in whatever format gen needs. Hopefully this will be clear in the tables that follow

6.4.1: Operations Within Expressions

In fact, we do a little more and generate code for assignment statements.

We will use two attributes code and address. For a parse tree node the code attribute gives the three address code to evaluate the input derived from that node. In particular, code at the root performs the entire assignment statement. there.

The attribute addr at a node is the address that holds the value calculated by the code at the node. Recall that unlike real code for a real machine our 3-address code doesn't reuse addresses.

As one would expect for expressions, all the attributes in the table to the right are synthesized. The table is for the expression part of the lab 3 grammar. To save space let's use as for assignment-statement, lv for lvalue, e for expression, t for term, and f for factor. Since we will be covering arrays a little later, we do not consider LET array-element.

6.4.2: Incremental Translation

We saw this in chapter 2.

The method in the previous section generates long strings and we walk the tree. By using SDT instead of using SDD, you can output parts of the string as each node is processed.

6.4.3: Addressing Array Elements

The idea is that you associate the base address with the array name. That is, the offset stored in the identifier table is the address of the first element of the array. The indices and the array bounds are used to compute the amount, often called the offset (unfortunately, we have already used that term), by which the address of the referenced element differs from the base address.

One Dimensional Arrays

For one dimensional arrays, this is especially easy: The address increment is the width of each element times the index (assuming indexes start at 0). So the address of A[i] is the base address of A plus i times the width of each element of A.

The width of each element is the width of what we have called the base type. So for an ID the element width is sizeof(getBaseType(ID.entry.type)). For convenience we define getBaseWidth by the formula

        getBaseWidth(ID.entry) = sizeof(getBaseType(ID.entry.type))

Two Dimensional Arrays

Let us assume row major ordering. That is, the first element stored is A[0,0], then A[0,1], ... A[0,k-1], then A[1,0], ... . Modern languages use row major ordering.

With the alternative column major ordering, after A[0,0] comes A[1,0], A[2,0], ... .

For two dimensional arrays the address of A[i,j] is the sum of three terms

  1. The base address of A.
  2. The distance from A to the start of row i. This is i times the width of a row, which is i times the number of elements in a row times the width of an element. The number of elements in a row is the column array bound.
  3. The distance from the start of row i to element A[i,j]. This is j times the width of an element.

(In some languages A[i,j] is written A[i][j].)

Higher Dimensional Arrays

The generalization to higher dimensional arrays is clear.

================ Start Lecture #11 ================

6.4.4: Translation of Array References

Translating Array References
ProductionSemantic Rules


as → lv = e ; as.code = e.code || lv.code || gen(*lv.addr = e.addr)


lv → ID lv.addr = new Temp()
lv.code = gen(lv.addr = &get(ID.lexeme))


lv → let ae lv.addr = ae.addr
lv.code = ae.code


ae → ID [ e ] ae.t1 = new Temp()
ae.t2 = new Temp()
ae.addr = new Temp()
ae.code = e.code || gen(ae.t1 = e.addr * getBaseWidth(ID.entry)) ||
      gen(ae.t2 = &get(ID.lexeme)) ||
      gen(ae.addr = ae.t2 + ae.t1)

Let's go over this carefully, especially the generated code and its use of addresses.

The book (both additions are the same in this respect) included a[i] as a legal address for three-address code. Last time, I did not appreciate the significance of this address form and thought it was just a convenience. In fact it is a special form.

Since the goal of the semantic rules is precisely generating such code, I could have used a[i]. I did not because
  1. Since we are restricted to one dimensional arrays, the full code generation for the address of an element is not hard and
  2. I thought it would be instructive to see the full address generation without hiding some of it under the covers.

It was definitely instructive for me! The rules for addresses in 3-address code also include

    a = &b
    a = *b
    *a = b
  
which are other special forms. They have the same meaning as in C.

I believe the SDD on the right if given a[3]=5, with a an integer array will generate

    t$1 = 3*4    // t$n are the temporary names from new TEMP()
    t$2 = &a
    t$3 = t$2 + t$1
    *t3 = 5
  

I also added an & to the non-array production lv→ID so that both could be handled by the same semantic rule for as→lv=e.

Homework: Write the SDD using the a[i] special form instead of the & and * special forms.

This is an exciting moment. At long last we can compile a full program!

Recall the program we could partially handle.

    procedure test () is
        y : integer;
        type t is array of real;
        x : t[10];
    begin
        y = 5;        // we haven't yet done statements
        x[2] = y;     // type error?
    end;
  
Now we can do the statements.

What about the possible type error?

  1. We could ignore errors.
  2. We could assume the intermediate language permits mismatched types. Final code generation would then need to generate conversion code or signal an error.
  3. We could change the program to use only one type.
  4. We could learn about type checking and conversions.

Let's take the last option.

Homework: What code is generated for the program written above?

6.5: Type Checking

Remark: We are back to chapter 6 in 1e.

Type Checking includes several aspects.

  1. The language comes with a type system, i.e., a set of rules saying what types can appear where.
  2. The compiler assigns a type expression to parts of the source program.
  3. The compiler checks that the type usage in the program conforms to the type system for the language.

All type checking could be done at run time: The compiler generates code to do the checks. Some languages have very weak typing; for example, variables can change their type during execution. Often these languages need run-time checks. Examples include lisp, snobol, apl.

A sound type system guarantees that all checks can be performed prior to execution. This does not mean that a given compiler will make all the necessary checks.

An implementation is strongly typed if compiled programs are guaranteed to run without type errors.

6.5.1: Rules for Type Checking

There are two forms of type checking.

  1. We will learn type synthesis where the types of parts are used to infer the type of the whole. For example, integer+real=real.
  2. Type inference is very slick. The type of a construct is determined from usage. This permits languages like ML to check types even though names need not be declared.

We consider type checking for expessions. Checking statements is very similar. View the statement as a function having its components as arguments and returning void.

6.5.2: Type Conversions

A very strict type system would do no automatic conversion. Instead it would offer functions for the programer to explicitly convert between selected types. Then either the program has compatible types or is in error.

However, we will consider a more liberal approach in which the language permits certain implicit conversions that the compiler is to supply. This is called type coercion. Explicit conversions supplied by the programmer are called casts. widening

We continue to work primarily with the two types used in lab 3, namely integer and real, and postulate a unary function denoted (real) that converts an integer into the real having the same value. Nonetheless, we do consider the more general case where there are multiple types some of which have coercions (often called widening). For example in C/Java, int can be widened to long, which in turn can be widened to float as shown in the figure to the right.

Mathematically the hierarchy on the right is a partially order set (poset) in which each pair of elements has a least upper bound (LUB). For many binary operators (all the arithmetic ones we are considering, but not exponentiation) the two operands are converted to the LUB. So adding a short to a char, requires both to be converted to an int. Adding a byte to a float, requires the byte to be converted to a float (the float remains a float and is not converted).

Checking and Coercing Types for Addition

The steps for addition, subtraction, multiplication, and division are all essentially the same: Convert each types if necessary to the LUB and then perform the arithmetic on the (converted or original) values. Note that conversion requires the generation of code.

Two functions are convenient.

  1. LUB(t1,t2) returns the type that is the LUB of the two given types. It signals an error if there is no LUB, for example if one of the types is an array.
  2. widen(a,t,w,newcode,newaddr). Given an address a of type t, and a (hopefully) wider address w, produce the instructions newcode needed so that the address newaddr is the conversion of address a to type w.

LUB is simple, just look at the address latice. If one of the type arguments is not in the lattice, signal an error; otherwise find the lowest common ancestor.

widen is more interesting. It involves n2 cases for n types. Many of these are error cases (e.g., if t wider than w). Below is the code for our situation with two possible types integer and real. The four cases consist of 2 nops (when t=w), one error (t=real; w=integer) and one conversion (t=integer; w=real).

    widen (a:addr, t:type, w:type, newcode:string, newaddr:addr)
      if t=w
        newcode = ""
        newaddr = a
      else if t=integer and w=real
        newaddr = new Temp()
        newcode = gen(newaddr = (real) a)
      else signal error
  

With these two functions it is not hard to modify the rules to catch type errors and perform coercions for arithmetic expressions.

  1. Maintain the type of each operand by defining type attributes for e, t, and f.
  2. Coerce each operand to the LUB.

This requires that we have type information for the base entities, identifiers and numbers. The lexer can supply the type of the numbers. We retrieve it via get(NUM.type).

It is more interesting for the identifiers. We insert that information when we process declarations. So we now have another semantic check: Is the identifier declared before it is used?

I will use the function get(ID.type), which returns the type from the identifier table and signals an error if it is not there. The original SDD for assignment statements was here and the changes for arrays was here.

Assignment Statements With Type Checks and Coercions
ProductionSemantic Rule


as → lv = e widen(e.addr, e.type, lv.type, as.code1, as.addr1)
as.code = lv.code || e.code || as.code1 || gen(*lv.addr = as.addr1)


lv → ID lv.addr = new TEMP()
lv.type = get(ID.type)
lv.code = gen(lv.addr = &get(ID.lexeme))


lv → let ae lv.addr = ae.addr
lv.type = ae.type
lv.code = ae.code


ae → ID [ e ] ae.type = getBaseType(ID.entry.type)
ae.t1 = new Temp()
ae.t2 = new Temp()
ae.addr = new Temp()
ae.code = e.code || gen(ae.t1 = e.addr * getBaseWidth(ID.entry)) ||
          gen(ae.t2 = &get(ID.lexeme)) ||
          gen(ae.addr = ae.t2 + ae.t1)


e → t e.addr = t.addr
e.type = t.type
e.code = t.code


e → e1 + t e.addr = new Temp()
e.type = LUB(e1.type, t.type)
widen(e1.addr, e1.type, e.type, e.code1, e.addr1)
widen(t.addr, t.type, e.type, e.code2, e.addr2)
e.code = e1.code || t.code || e.code1 || e.code2 || gen(e.addr = e.addr1 + e.addr2)


e → e1 - t e.addr = new Temp()
e.type = LUB(e1.type, t.type)
widen(e1.addr, e1.type, e.type, e.code1, e.addr1)
widen(t.addr, t.type, e.type, e.code2, e.addr2)
e.code = e1.code || t.code || e.code1 || e.code2 || gen(e.addr = e.addr1 - e.addr2)


t → f t.addr = f.addr
t.type = f.type
t.code = f.code


t → t1 * f t.addr = new Temp()
t.type = LUB(t1.type, f.type)
widen(t1.addr, t1.type, t.type, t.code1, t.addr1)
widen(f.addr, f.type, t.type, t.code2, t.addr2)
t.code = t1.code || f.code || t.code1 || t.code2 || gen(t.addr = t.addr1 * t.addr2)


t → t1 / f t.addr = new Temp()
t.type = LUB(t1.type, f.type)
widen(t1.addr, t1.type, t.type, t.code1, t.addr1)
widen(f.addr, f.type, t.type, t.code2, t.addr2)
t.code = t1.code || f.code || t.code1 || t.code2 || gen(t.addr = t.addr1 / t.addr2)


f → ( e ) f.addr = e.addr
f.type = e.type
f.code = e.code


f → ID f.addr = get(ID.lexeme)
f.type = get(ID.type)
f.code = ""


f → NUM f.addr = get(NUM.lexeme)
f.type = get(NUM.type)
f.code = ""

Homework: Same question as the previous homework (What code is generated for the program written above?). But the answer is different!

6.5.3: Overloading of Functions and Operators

Skipped.

Overloading is when a function or operator has several definitions depending on the types of the operands and result.

6.5.4: Type Inference and Polymorphic Functions

Skipped.

6.5.5: An Algorithm for Unification

Skipped.

6.6: Control Flow

Remark: Section 8.4 in 1e.

Control flow includes the study of Boolean expressions, which have two roles.

  1. They can be computed and treated similar to integers or real. Once can declare Boolean variables, there are boolean constants and boolean operators. There are also relational operators that produce Boolean values from arithmetic operands. From this point of view, Boolean expressions are similar to the expressions we have already treated. Our previous semantic rules could be modified to generate the code needed to evaluate these expressions.
  2. They are used in certain statements that alter the normal flow of control. In this regard, we have something new to learn.

6.6.1: Boolean Expressions

One question that comes up with Boolean expressions is whether both operands need be evaluated. If we need to evaluate A or B and find that A is true, must we evaluate B? For example, consider evaluating

     A=0 OR  3/A < 1.2
  
when A is zero.

This comes up some times in arithmetic as well. Consider A*F(x). If the compiler knows that for this run A is zero must it evaluate F(x)? Don't forget that functions can have side effects,

6.6.2: Short-Circuit Code

This is also called jumping code. Here the Boolean operators AND, OR, and NOT do not appear in the generated instruction stream. Instead we just generate jumps to either the true branch or the false branch flow of control

6.6.3: Flow-of-Control Statements

This time I will follow 2e and use C/Java grammar rather than lab 3 grammar since lab 3 is basically a subset.

So our grammar is (S for statement, B for boolean expression)

  S → if ( B ) S1
  S → if ( B ) S1 else S2
  S → while ( B ) S1
What is missing from lab 3 is the elseless if and Boolean operators.

The idea is simple.

  1. In this section we will produce an SDD for these three compound statements under the assumption that the SDD for B generates jumps to the labels B.true and B.false (depending of course on whether B is true or false).
  2. In the next section we give the needed SDD for B.
  3. I don't know why the sections aren't in the reverse order and I came close to reversing the order of presentation.
  4. The diagrams on the right give the idea.
  5. The table below gives the details.

If and While SDDs
ProductionSemantic RulesKind



P → SS.next = newlabel()Inherited
P.code = S.code || label(S.next)Synthesized



S → if ( B ) S1 B.true = newlabel()Inherited
B.false = S.nextInherited
S1.next = S.nextInherited
S.code = B.code || label(B.true) || S1.codeSynthesized



S → if ( B ) S1 else S2 B.true = newlabel()Inherited
B.false = newlabel()Inherited
S1.next = S.nextInherited
S2.next = S.nextInherited
S.code = B.code || label(B.true) || S1.code
      || gen(goto S.next) || label(B.false) || S2.code
Synthesized



S → while ( B ) S1 begin = newlabel()Synthesized
B.true = newlabel()Inherited
B.false = S.nextInherited
S1.next = beginInherited
S.code = label(begin) || B.code || label(B.true) || S1.code || gen(goto begin) Synthesized



S → S1 S2 S1.next = newlabel()Inherited
S2.next = S.nextInherited
S.code = S1.code || label(S1.next) || S2.code Synthesized

Homework: Give the SDD for a repeat statement
Repeat S while B

6.6.4: Control-Flow Translation of Boolean Expressions

Boolean Expressions
ProductionSemantic RulesKind



B → B1 || B2 B1.true = B.trueInherited
B1.false = newlabel()Inherited
B2.true = B.trueInherited
B2.false = B.falseInherited
B.code = B1.code || label(B1.false) || B2.code Synthesized



B → B1 && B2 B1.true = newlabel()inherited
B1.false = B.falseinherited
B2.true = B.trueinherited
B2.false = B.falseinherited
B.code = B1.code || label(B1.true) || B2.code Synthesized



B → ! B1 B1.true = B.falseInherited
B1.false = B.trueInherited
B.code = B1.codeSynthesized



B → E1 relop E2 B.code = E1.code || E2.code
      || gen(if E1.addr relop.lexeme E2.addr goto B.true)
      || gen(goto B.false)
Synthesized



B → trueB.code = gen(goto B.true)Synthesized



B → falseB.code = gen(goto B.false)Synthesized



B → IDB.code = gen(if get(ID.lexeme) goto B.true)
      || gen(goto B.false)
Synthesized

Do on the board the translation of

    if ( x < 5 || x > 10 && x == y ) x = 3 ;
  

We get

        if x < 5 goto L2
        goto L3
    L3: if x > 10 goto L4
	goto L1
    L4: if x == y goto L2
	goto L1
    L2: x = 3
  

Note that there are three extra gotos. One is a goto the next statement. Two others could be eliminated by using ifFalse.

================ Start Lecture #12 ================

6.6.5: Avoiding Redundant Gotos

Skipped.

6.6.6: Boolean Values and Jumping Code

Remark: As mentioned before 6.6 in the notes is 6.6 in 2e and 8.4 in 1e. However the third level material is not is the same order. In particular this section (6.6.6) is very early in 8.4.

If ther are boolean variables (or variables into which a boolean value can be placed), we can have boolean assignment statements. That is we might evaluate boolean expressions outside of control flow statements.

Recall that the code we generated for boolean expressions (inside control flow statements) used inherited attributes to push down the tree the exit labels B.true and B.false. How are we to deal with Boolean assignment statements?

Two Methods for Booleans: Method 1

Up to now we have used the so called jumping code method for Boolean quantities. We evaluated Boolean expressions (in the context of control flow statements) by using inherited attributes to push down the tree the true and false exits (i.e., the target locations to jump to if the expression evaluates to true and false).

With this method if we have a Boolean assignment statement, we just let the true and false exits lead to statements

    LHS = true
    LHS = false
  
respectively.

Two Methods for Booleans: Method 2

In the second method we simply treat boolean expressions as expressions. That is, we just mimic the actions we did for integer/real evaluations. Thus Boolean assignment statements like
a = b OR (c AND d AND (x < y))
just work.

For control flow statements like

    while (boolean-expression) statement-list end
    if (boolean-expression) statement-list else statement-list end
  
we simply evaluate the boolean expression as if it was part of an assignment statement and then have two jumps to where we should go if the result is true or false.

However this is wrong. In C if (a=0 || 1/a > f(a)) is guaranteed not to divide by zero and the above implementation fails to provide this guarantee. We must somehow implement short-circuit boolean evaluation.

6.7: Backpatching

Skipped.

Our intermediate code uses symbolic labels. At some point these must be translated into addresses of instructions. If we use quads all instructions are the same length so the address is just the number of the instruction. Sometimes we generate the jump before we generate the target so we can't put in the instruction number on the fly. Indeed, that is why we used symbolic labels. The easiest method of fixing this up is to make an extra pass (or two) over the quads to determine the correct instruction number and use that to replace the symbolic label. This is extra work; a more efficient technique, which is independent of compilation, is called backpatching.

6.8: Switch Statements

Evaluate an expression, compare it with a vector of constants that are viewed as labels of the arms of the switch, and execute the matching arm (or a default).

The C language is unusual in that the various cases are just labels for a giant computed goto at the beginning. The more traditional idea is that you execute just one of the arms, as in a series of

    if
    else if
    else if
    ...
    end if
  

6.8.1: Translation of Switch-Statements

  1. Simplest implementation to understand is to just transform the switch into the series if else if's above. This executes roughly 2k jumps (worst case) for k cases.
  2. Instead you can begin with jumps to each case. This executes roughly k jumps.
  3. Create a jump table. If the constant values lie in a small range and are dense, then make a list of jumps one for each number in the range and use the value computed to determine which of these jumps to jump to. This executes 2 jumps.

6.8.2: Syntax-Directed Translation of Switch-Statements

The lab 3 grammar does not have a switch statement so we won't do a detailed SDD.

Such an SDD would be organized as follows.

  1. When you process the switch (E) ... production, call newlabel() to generate labels for next and test which are put into inherited and synthesized attributes respectively.

  2. Then the expression is evaluated with the code and the address synthesized up.

  3. The code for the switch has after the code for E a goto test.

  4. Each case begins with a newlabel(). The code for the case begins with this label and then the translation of the arm itself and ends with a goto next. The generated label paired with the value for this case is added to an inherited attribute representing a queue of these pairs (actually this is done by some production like
          cases → case cases | ε
    As usual the queue is sent back up the tree by the epsilon production.

  5. When we get to the end of the cases we are back at the switch production which now adds code to the end. Specifically, the test label is gen'ed and then a series of
          if E.addr = Vi goto Li
    statements, where each Li,Vi pair is from the generated queue.

6.9 Intermediate Code for Procedures

Much of the work for procedures involves storage issues and the run time environment; this is discussed in the next chapter.

In order to support inter-procedural type checking by the compiler, we need to define the called procedure in the calling procedure, which the lab 3 grammar doesn't support except for calling itself recursively. So the best we can do is type check recursive calls.

The basic scheme for type checking recursive (or other calls) is to generate a table entry for the procedure that contains its signature, i.e., the types of its parameters and its result type.

Recall the SDD for declarations. These semantic rules pass up the totalSize to the
      ds → d ds
production.

What is needed is for the ps (parameters) to do an analogous thing with their declarations but also (or perhaps instead) pass up a representation of the declarations themselves which when it reaches the top is the signature for a procedure and when put together with the return is the signature for a function.

More serious is supporting nested procedure definitions (defining a procedure inside a procedure). Lab 3 doesn't support this because a procedure or function definition is not a declaration. It would be easy to enhance the grammar to fix this, but the serious work is that then you need nested identifier tables.

Our lexer doesn't support this. So you would remove table building from the lexer and instead do it in the parser and when a new scope (procedure definition, record definition, begin block) arises you push the current tables on a stack and begin a new one. When the nested scope ends, you pop the tables.

Chapter 7: Run Time Environments

Homework: Read Chapter 7.

7.1: Storage Organization

We are discussing storage organization from the point of view of the compiler, which must allocate space for programs to be run. In particular, we are concerned with only virtual addresses and treat them uniformly.

This should be compared with an operating systems treatment, where we worry about how to effectively map this configuration to real memory. For example see see these two diagrams in my OS class notes, which illustrate an OS difficulty with our allocation method, which uses a very large virtual address range and one solution.

Some system require various alignment constraints. For example 4-byte integers might need to begin at a byte address that is a multiple of four. Unaligned data might be illegal or might lower performance. To achieve proper alignment padding is often used.

Areas (segments) of Memory

runtime memory
  1. The code (often called text in OS-speak) is fixed size and unchanging (self-modifying code is long out of fashion). If there is OS support it could be marked execute only (or perhaps read and execute, but not write). All other areas would be marked non-executable (except for systems like lisp that execute their data).
  2. There is likely data of fixed size whose need can be determined by the compiler by examining the program's structure (and not by determining the program's execution pattern). One example is global data. Storage for this data would be allocated in the next area right after the code. A key point is that since the code and this area are of fixed size that does not change during execution, they, unlike the next two areas, have no need for an expansion region.
  3. The stack is used for memory whose lifetime is stack-like. It is organized into activation records that are created as a procedure is called and destroyed when the function exits. It abuts the area of unused memory so can grow easily. Typically the stack is stored at the highest virtual addresses and grows downward (toward small addresses). However, it is sometimes easier in describing the activation records and their uses to pretend that the addresses are increasing (so that increments are positive).
  4. The heap is used for data whose lifetime is not as easily described. This data is allocated by the program itself, typically either with a language construct, such as new, or via a library function call, such as malloc(). It is deallocated either by another executable statement, such as a call to free(), or automatically by the system.

7.1.1: Static Versus Dynamic Storage Allocation

Much (often most) data cannot be statically allocated. Either its size is not know at compile time or its lifetime is only a subset of the program's execution.

Early versions of Fortran used only statically allocated data. This required that each array had a constant size specified in the program. Another consequence of supporting only static allocation was that recursion was forbidden (otherwise the compiler could not tell how many versions of a variable would be needed).

Modern languages, including newer versions of Fortran, support both static and dynamic allocation of memory.

The advantage supporting dynamic storage allocation is the increased flexibility and storage efficiency possible (instead of declaring an array to have a size adequate for the largest data set; just allocate what is needed). The advantage of static storage allocation is that it avoids the runtime costs for allocation/deallocation and may permit faster code sequences for referencing the data.

An (unfortunately, all too common) error is a so-called memory leak where a long running program repeated allocates memory that it fails to delete, even after it can no longer be referenced. To avoid memory leaks and ease programming, several programming language systems employ automatic garbage collection. That means the runtime system itself can determine if data can no longer be referenced and if so automatically deallocates it.

7.2: Stack Allocation of Space

Achievements.

  1. Space shared by procedure calls that have disjoint durations (despite being unable to check disjointness statically).
  2. The relative address of each nonlocal variable is constant throughout execution.

7.2.1: Activation Trees

Recall the fibonacci sequence 1,1,2,3,5,8, ... defined by f(1)=f(2)=1 and, for n>2, f(n)=f(n-1)+f(n-2). Consider the function calls that result from a main program calling f(5). On the left we show the calls and returns linearly and on the right in tree form. The latter is sometimes called the activation tree or call tree.

activation tree
    System starts main
        enter f(5)
            enter f(4)
                enter f(3)
		    enter f(2)
		    exit f(2)
		    enter f(1)
		    exit f(1)
                exit f(3)
                enter f(2)	       int a[10];
                exit f(2)	       int main(){
            exit f(4)		           int i;
            enter f(3)		           for (i=0; i<10; i++){
	        enter f(2)	               a[i] = f(i);
		exit f(2)	           }
		enter f(1)	       }
		exit f(1)	       int f (int n) {
            exit f(3)		           if (n<3)  return 1;
        exit f(5)		           return f(n-1)+f(n-2);
    main ends			       }
  

We can make the following observation about these procedure calls.

  1. If an activation of p calls q, then that activation of q terminates no later than the activation of p.
  2. The order of activations (procedure calls) corresponds to a preorder traversal of the call tree.
  3. The order of de-activations (procedure returns) corresponds to postorder traversal of the call tree.
  4. If execution is currently in an activation corresponding to a node N of the activation tree, then the activations that are currently live are those corresponding to N and its ancestors in the tree. They were called in the order given by the root-to-N path in the tree and the returns will occur in the reverse order.
activation record

7.2.2: Activation Records (ARs)

The information needed for each invocation of a procedure is kept in a runtime data structure called an activation record (AR) or frame. The frames are kept in a stack called the control stack.

At any point in time the number of frames on the stack is the current depth of procedure calls. For example, in the fibonacci execution shown above when f(4) is active there are three activation records on the control stack.

ARs vary with the language and compiler implementation. Typical components are described and pictured below. In the diagrams the stack grows down the page.

  1. Temporaries. For example, recall the temporaries generated during expression evaluation. Often these can be held in machine registers, when not possible the temporary area is used.
  2. Data local to the procedure being activated.
  3. Saved status from the caller, which typically includes the return address and the machine registers. The register values are restored when control returns to the caller.
  4. The access link is described below.
  5. The control link connects the ARs by pointing to the AR of the caller.
  6. Returned values are normally (but not always) placed in registers.
  7. The first few parameters are normally (but not always) placed in registers.
control stack

The diagram on the right shows (part of) the control stack for the fibonacci example at three points during the execution. In the upper left we have the initial state, We show the global variable a, although it is not in an activation record and actually is allocated before the program begins execution (it is statically allocated; recall that the stack and heap are each dynamically allocated). Also shown is the activation record for main, which contains storage for the local variable i.

Below the initial state we see the next state when main has called f(1) and there are two activation records, one for main and one for f. The activation record for f contains space for the parameter n and and also for the result. There are no local variables in f.

At the far right is a later state in the execution when f(4) has been called by main and has in turn called f(2). There are three activation records, one for main and two for f. It is these multiple activations for f that permits the recursive execution. There are two locations for n and two for the result.

7.2.3: Calling Sequences

The calling sequence, executed when one procedure (the caller) calls another (the callee), allocates an activation record (AR) on the stack and fills in the fields. Part of this work is done by the caller; the remainder by the callee. Although the work is shared, the AR is called the callee's AR.

Since the procedure being called is defined in one place, but called from many, there are more instances of the caller activation code than of the callee activation code. Thus it is wise, all else being equal, to assign as much of the work to the callee.

creating ARs
  1. Values computed by the caller are placed before any items of size unknown by the caller. This way they can be referenced by the caller using fixed offsets. One possibility is to place values computed by the caller at the beginning of the activation record (AR), i.e., near the AR of the caller. The number of arguments may not be the same for different calls of the same function (so called varargs, e.g. printf() in C).
  2. Fixed length items are placed next. These include the links and the saved status.
  3. Finally come items allocated by the callee whose size is known only at run-time, e.g., arrays whose size depends on the parameters.
  4. The stack pointer sp is between the last two so the temporaries and local data are actually above the stack. This would seem more surprising if I used the book's terminology, which is top_sp. Fixed length data can be referenced by fixed offsets (known to the intermediate code generator) from the sp.

The top picture illustrates the situation where a pink procedure (the caller) calls a blue procedure (the callee). Also shown is Blue's AR. Note that responsibility for this single AR is shared by both procedures. The picture is just an approximation: For example, the returned value is actually the Blue's responsibility (although the space might well be allocated by Pink. Also some of the saved status, e.g., the old sp, is saved by Pink.

The bottom picture shows what happens when Blue, the callee, itself calls a green procedure and thus Blue is also a caller. You can see that Blue's responsibility includes part of its AR as well as part of Green's. creating ars 2

Calling Sequence

  1. The caller evaluates the arguments. (I use arguments for the caller, parameters for the callee.)
  2. The caller stores the return address and the (soon-to-be-updated) sp in the callee's AR.
  3. The caller increments sp so that instead of pointing into its AR, it points to the corresponding point in the callee's AR.
  4. The callee saves the registers and other (system dependent) information.
  5. The callee allocates and initializes its local data.
  6. The callee begins execution.

Return Sequence

  1. The callee stores the return value near the parameters. Note that this address can be determined by the caller using the old (soon-to-be-restored) sp.
  2. The callee restores sp and the registers.
  3. The callee jumps to the return address.

Note that varagrs are supported. var size stack

7.2.4: Variable-Length Data on the Stack

There are two flavors of variable-length data.

It is the second flavor that we wish to allocate on the stack. The goal is for the (called) procedure to be able to access these arrays using addresses determinable at compile time even though the size of the arrays (and hence the location of all but the first) is not know until the program is called and indeed often differs from one call to the next.

The solution is to leave room for pointers to the arrays in the AR. These are fixed size and can thus be accessed using static offsets. Then when the procedure is invoked and the sizes are known, the pointers are filled in and the space allocated.

A small change caused by storing these variable size items on the stack is that it no longer is obvious where the real top of the stack is located relative to sp. Consequently another pointer (call it real-top-of-stack) is also kept. This is used on a call to tell where the new allocation record should begin.

7.3: Access to Nonlocal Data on the Stack

As we shall see the ability of procedure p to access data declared outside of p (either declared globally outside of all procedures or declared inside another procedure q) offers interesting challenges.

7.3.1: Data Access Without Nested Procedures

In languages like standard C without nested procedures, visible names are either local to the procedure in question or are declared globally.

  1. For global names the address is known statically at compile time providing there is only one source file. If multiple source files, the linker knows. In either case no reference to the activation record is needed; the addresses are know prior to execution.
  2. For names local to the current procedure, the address needed is in the AR at a known-at-compile-time constant offset from the sp. In the case of variable size arrays, the constant offset refers to a pointer to the actual storage.

7.3.2: Issues With Nested Procedures

With nested procedures a complication arises. Say g is nested inside f. So g can refer to names declared in f. These names refer to objects in the AR for f; the difficulty is finding that AR when g is executing. We can't tell at compile time where the (most recent) AR for f will be relative to the current AR for g since a dynamically-determined number of routines could have been called in the middle.

There is an example in the next section. in which g refers to x, which is declared in the immediately outer scope (main) but the AR is 2 away because f was invoked in between. (In that example you can tell at compile time what was called in what order, but with a more complicated program having data-dependent branches, it is not possible.)

7.3.3: A language with Nested Procedure Declarations

As we have discussed, the 1e, which you have, uses pascal, which many of you don't know. The 2e, which you don't have uses C, which you do know.

Since pascal supports nested procedures, this is what the 1e uses to give examples.

The 2e asserts (correctly) that C doesn't have nested procedures so introduces ML, which does (and is quite slick), but which unfortunately many of you don't know and I haven't used. Fortunately a common extension to C is to permit nested procedures. In particular, gcc supports nested procedures. To check my memory I compiled and ran the following program.

#include <stdio.h>

int main (int argc, char *argv[])
{
    int x = 10;

    void g(int y)
    {
        int z = x;
	return;
    }

    int f (int y)
    {
	g(y);
	return y+1;
    }

    printf("The answer is %d\n", f(x));
    return 0;
}

The program compiles without errors and the correct answer of 11 is printed.

So we can use C (really the GCC, et al extension of C).

7.3.4: Nesting Depth

Outermost procedures have nesting depth 1. Other procedures have nesting depth 1 more than the nesting depth of the immediately outer procedure. In the example above main has nesting depth 1; both f and g have nesting depth 2.

7.3.5: Access Links

The AR for a nested procedure contains an access link that points to the AR of the (most recent activation of the immediately outer procedure). So in the example above the access link for all activations of f and g would point to the AR of the (only) activation of main. Then for a procedure P to access a name defined in the 3-outer scope, i.e., the unique outer scope whose nesting depth is 3 less than that of P, you follow the access links three times.

The question is how are the access links maintained.

7.3.6: Manipulating Access Links

Let's assume there are no procedure parameters. We are also assuming that the entire program is compiled at once. For multiple files the main issues involve the linker, which is not covered in this course. I do cover it a little in the OS course.

Without procedure parameters, the compiler knows the name of the called procedure and, since we are assuming the entire program is compiled at once, knows the nesting depth.

Let the caller be procedure R (the last letter in caller) and let the called procedure be D. Let N(f) be the nesting depth of f. I did not like the presentation in 2e (which had three cases and I think did not cover the example above). I made up my own and noticed it is much closer to 1e (but makes clear the direct recursion case, which is explained in 2e). I am surprised to see a regression from 1e to 2e, so make sure I have not missed something in the cases below.

  1. N(D)>N(R). The only possibility is for D to be immediately declared inside R. Then when compiling the call from R to D it is easy to include code to have the access link of D point to the AR of R.
  2. N(D)≤N(R). This includes the case D=R, i.e., a direct recursive call. For D to be in the scope of R, there must be another procedure P enclosing both D and R, with D immediately inside P, i.e., N(D)=N(P)+1 and N(R)=N(P)+1+k, with k≥0.
          P() {
            D() {...}
            P1() {
              P2() {
               ...
                    Pk() {
                      R(){... D(); ...}
                    }
               ...
              }
            }
          }
        
    Our goal while creating the AR for D at the call from R is to set the access link to point to the AR for P. Note that this entire structure in the skeleton code shown is visible to the compiler. Thus, the current (at the time of the call) AR is the one for R and if we follow the access links k+1 times we get a pointer to the AR for P, which we can then place in the access link for the being-created AR for D.

When k=0 we get the gcc code I showed before and also the case of direct recursion where D=R.

7.3.7: Access Links for Procedure Parameters

Basically skipped. The problem is that, if f calls g giving with a parameter of h (or a pointer to h in C-speak) and the g calls this parameter (i.e., calls h), g might not know the context of h. The solution is for f to pass to g the pair (h, the access link of h) instead of just passing h. Naturally, this is done by the compiler, the programmer is unaware of access links.

7.3.8: Displays

Basically skipped. In theory access links can form long chains (in practice nesting depth rarely exceeds a dozen or so). A display is an array in which entry i points to the most recent (highest on the stack) AR of depth i.

7.4: Heap Management

Almost all of this section is covered in the OS class.

7.4.1: The Memory Manager

Covered in OS.

7.4.2: The Memory Hierarchy of a Computer

Covered in Architecture.

7.4.3: Locality in Programs

Covered in OS.

7.4.4: Reducing (external) Fragmentation

Covered in OS.

7.4.5: Manual Deallocation Requests

Stack data is automatically deallocated when the defining procedure returns. What should we do with heap data explicated allocated with new/malloc?

The manual method is to require that the programmer explicitly deallocate these data. Two problems arise.

  1. Memory leaks. The programmer forgets to deallocate.
    	loop
    	   allocate X
    	   use X
    	   forget to deallocate X
          
    As this program continues to run it will require more and more storage even though is actual usage is not increasing significantly.
  2. Dangling References. The programmer forgets that they did a deallocate.
    	allocate X
    	use X
    	deallocate X
    	100,000 lines of code not using X
    	use X
          

Both can be disastrous.

7.5: Introduction to Garbage Collection

The system detects data that cannot be accessed (no direct or indirect references exist) and deallocates the data automatically.

Covered in programming languages???

Skipped

7.5.1: Design Goals for Garbage Collectors

Skipped

7.5.2: Reachability

Skipped.

7.5.3: Reference Counting Garbage Collectors

Skipped.

7.6: Introduction to Trace-Based Collection

7.6.1: A Basic Mark-and-Sweep Collector

Skipped.

7.6.2:Basic Abstraction

Skipped.

7.6.3: Optimizing Mark-and-Sweep

Skipped.

7.6.4: Mark-and-Compact Garbage Collectors

Skipped.

7.6.5: Copying Collectors

Skipped.

7.6.6: Comparing Costs

Skipped.

7.7: Short Pause Garbage Collection

7.7.1: Incremental Garbage Collection

Skipped.

7.7.2: Incremental Reachability Analysis

Skipped.

7.7.3: Partial Collection Basics

Skipped.

7.7.4: Generational Garbage Collection

Skipped.

7.7.5: The Train Algorithm

Skipped.

7.8: Advanced Topics in Garbage Collection

Skipped

7.8.1: Parallel and Concurrent Garbage Collection

Skipped.

7.8.2: Partial Object Relocation

Skipped.

7.8.3: Conservative Collection for Unsafe Languages

Skipped.

7.8.4: Weak References

Skipped.

================ Start Lecture #13 ================

Chapter 8: Code Generation

Remark: This is chapter 9 in 1e.

Homework: Read Chapter 8.

Goal: Transform intermediate code + tables into final machine (or assembly) code. Code generation + Optimization is the back end of the compoiler.

8.1: Issues in the Design of a Code Generator

8.1.1: Input to the Code Generator

As expected the input is the output of the intermediate code generator. We assume that all syntactic and semantic error checks have been done by the front end. Also all needed type conversions are already done and any type errors have been detected.

We are using three address instructions for our intermediate language. The these instructions have several representations, quads, triples, indirect triples, etc. In this chapter I will tend to write quads (for brevity) when I should write three-address instructions.

8.1.2: the Target Program

A RISC (Reduced Instruction Set Computer), e.g. PowerPC, Sparc, MIPS (popular for embedded systems), is characterized by

A CISC (Complex Instruct Set Computer), e.g. x86, x86-64/amd64 is characterized by

A stack-based computer is characterized by

  1. No registers
  2. Zero address instructions (operands/results implicitly on the stack)
  3. Top portion of stack kept in hidden registers

A Little History

IBM 701/704/709/7090/7094 (Moon shot, MIT CTSS) were accumulator based.

Stack based machines were believed to be good compiler targets. They became very unpopular when it was believed that register architecture would perform better. Better compilation (code generation) techniques appeared that could take advantage of the multiple registers.

Pascal P-code and Java byte-code are the machine instructions for a hypothetical stack-based machine, the JVM (Java Virtual Machine) in the case of Java. This code can be interpreted, or compiled to native code.

RISC became all the rage in the 1980s.

CISC made a gigantic comeback in the 90s with the intel pentium pro. A key idea of the pentium pro is that the hardware would dynamically translate a complex x86 instruction into a series of simpler RISC-like instructions called ROPs (RISC ops). The actual execution engine dealt with ROPs. The jargon would be that, while the architecture (the ISA) remained the x86, the micro-architecture was quite different and more like the micro-architecture seen in previous RISC processors.

Assemblers and Linkers

For maximum compilation speed, the compiler accepts the entire program at once and produces code that can be loaded and executed (the compilation system can include a simple loader and can start the compiled program). This was popular for student jobs when computer time was expensive. The alternative, where each procedure can be compiled separately, requires a linkage editor.

It eases the compiler's task to produce assembly code instead of machine code and we will do so. This decision increased the total compilation time since it requires an extra assembler pass (or two).

8.1.3: Instruction Selection

A big question is the level of code quality we seek to attain. For example can we simply translate one quadruple at a time. The quad
        x = y + z
can always (assuming x, y, and z are statically allocated, i.e., their address is a compile time constant off the sp) be compiled into

    LD  R0, y
    ADD R0, R0, z
    ST  x, R0
  
But if we apply this to each quad separately (i.e., as a separate. problem) then
        a = b + c
        d = a + e
is compiled into
    LD  R0, b
    ADD R0, R0, c
    ST  a, R0
    LD  R0, a
    ADD R0, e
    ST  d, R0
  
The fourth statement is clearly not needed since we are loading into R0 the same value that it contains. The inefficiency is caused by our compiling the second quad with no knowledge of how we compiled the first quad.

8.1.4: Register Allocation

Since registers are the fastest memory in the computer, the ideal solution is to store all values in registers. However, there are normally not nearly enough registers for this to be possible. So we must choose which values are in the registers at any given time.

Actually this problem has two parts.

  1. Which values should be stored in registers?
  2. Which register should each selected value be stored in

The reason for the second problem is that often there are register requirements, e.g., floating-point values in floating-point registers and certain requirements for even-odd register pairs (e.g., 0&1 but not 1&2) for multiplication/division.

8.1.5: Evaluation Order

Sometimes better code results if the quads are reordered. One example occurs with modern processors that can execute multiple instructions concurrently, providing certain restrictions are met (the obvious one is that the input operands must already be evaluated).

8.2: The Target Language

This is a delicate compromise between RISC and CISC. The goal is to be simple but to permit the study of nontrivial addressing modes and the corresponding optimizations. A charging scheme is instituted to reflect that complex addressing modes are not free.

8.2.1: A Simple Target Machine Model

We postulate the following (RISC-like) instruction set

  1. Load. LD dest, addr loads the register dest with the contents of the address addr.

    LD reg1, reg2 is a register copy.

    A question is whether dest can be a memory location or whether it must be a register. This is part of the RISC/CISC debate. In CISC parlance, no distinction is made between load and store, both are examples of the general move instruction that can have an arbitrary source and an arbitrary destination.

    We will normally not use a memory location for the destination of a load (or the target of a store). That is we do not permit memory to memory copy in one instruction.

    As will be see below we charge more for a memory location than for a register.

  2. Store. ST addr, src stores the value of the source src (register) into the address addr.

  3. Computation. OP dest, src1, src2 performs the operation OP on the two source operands src1 and src2. For RISC the three operands must be registers. If the destination is one of the sources the source is read first and then overwritten (using a master-slave flip-flop if it is a register).

  4. Unconditional branch. BR L transfers control to the (instruction with) label L.

  5. Conditional Branch. Bcond r, L transfers to the label L if register r satisfies the condition cond. For example,

          BNEG R0, joe

    branches to joe if R0 is negative.

The addressing modes are not RISC-like at first glance, as they permit memory locations to be operands. Again, note that we shall charge more for these.

  1. Variable name. This is shorthand (or assembler-speak) for the memory location containing x, i.e., the l-value of x.
  2. Indexed address. The address a(r), where a is a variable name and r is a register (number) specifies the address that is, the value-in-r bytes past the address specified by a.

    LD r1, a(r2) sets (the contents of R1) equal to

        contents(a+contents(r2)) NOT
        contents(contents(a)+contents(r2))

    That is, the l-value of a and not the r-value is used.

    If permitted outside a load or store instruction, this addressing mode would plant the CISC flag firmly in the ground.

  3. Indexed constant. An integer constant can be indexed by a register. So
        LD r1, 8(r4)
    sets r1 equal to contents(8+contents(r4))

  4. Indirect addressing. If I is an explicit integer constant and r is a register (number), the previous addressing mode tells us that I(r) refers to the address I+contents(r). The new addressing mode *I(r) refers to the address contents(I+contents(r)). The address *r is shorthand for *0(r). So
          LD r1, *r2
    sets the (contents of) r1 equal to (get ready)
          contents(contents(contents(r2)))
    and
          LD r1, *50(r2)
    sets r1 equal to
          contents(contents(50+contents(r2))).

  5. Immediate constant. If a constant is preceded by a # it is treated as an r-value instead of an l-value. So
          ADD r2, r2, #1
    is an increment instruction.

For many quads the standard naive (RISC-like) translation is 4 instructions.

  1. Load first source.
  2. Load second source.
  3. Do operation.
  4. Store result

Array assignment statements are also four instructions. We can't do A[i]=B[j] because that needs four addresses.

The instruction x=A[i] becomes (assuming each element of A is 4 bytes)

    LD  R0, i
    MUL R0, R0, #4
    LD  R0, A(R0)
    ST  x, R0
  

Similarly A[i]=x becomes

    LD  R0, x
    LD  R1, i
    MUL R1, R1, #4
    ST  A(R1), R0
  

The pointer reference x = *p becomes

    LD  R0, p
    LD  R0, 0(R0)
    ST  x, R0
  

The assignment through a pointer *p = x becomes

    LD  R0, x
    LD  R1, p
    ST  0(R1), R0
  

Finally if x < y goto L becomes

    LD   R0, x
    LD   R1, y
    SUB  R0, R0, R1
    BNEG R0, L
  

8.2.2: Program and Instruction Costs

The run-time cost of a program depends on (among other factors)

Here we just determine the first cost, and use quite a simple metric. We charge for each instruction one plus the cost of each addressing mode used.

Addressing modes using just registers have zero cost, while those involving memory addresses or constants are charged one. This corresponds to the size of the instruction since a memory address or a constant is assumed to be stored in a word right after the instruction word itself.

You might think that we are measuring the memory (or space) cost of the program not the time cost, but this is mistaken: The primary space cost is the size of the data, not the size of the instructions. One might say we are charging for the pressure on the I-cache.

For example, LD R0, *50(R2) costs 2, the additional cost is for the constant 50.

Homework: 9.1, 9.3

8.3: Address in the Target Code

There are 4 possibilities for addresses that must be generated depending on which of the following areas the address refers to.

  1. The text or code area. The size of this area is statically determined.
  2. The static area holding global constants. The size of this area is statically determined.
  3. The stack holding activation records. The size of this area is not known at compile time.
  4. The heap. The size of this area is not known at compile time.

8.3.1: Static Allocation

Returning to the glory days of Fortran, we consider a system with static allocation. Remember, that with static allocation we know before execution where all the data will be stored. There are no recursive procedures; indeed, there is no run-time stack of activation records. Instead the ARs are statically allocated by the compiler.

Caller Calling Callee

In this simplified situation, calling a parameterless procedure just uses static addresses and can be implemented by two instructions. Specifically,
      call procA
can be implemented by

    ST  callee.staticArea, #here+20
    BR  callee.codeArea
  

We are assuming, for convenience, that the return address is the first location in the activation record (in general it would be a fixed offset from the beginning of the AR). We use the attribute staticArea for the address of the AR for the given procedure (remember again that there is no stack and heap).

What is the mysterious #here+20?

The # we know signifies an immediate constant. We use here to represent the address of the current instruction (the compiler knows this value since we are assuming that the entire program, i.e., all procedures, are compiled at once). The two instructions listed contain 3 constants, which means that the entire sequence takes 5 words or 20 bytes. Thus here+20 is the address of the instruction after the BR, which is indeed the return address.

Callee Returning

With static allocation, the compiler knows the address of the the AR for the callee and we are assuming that the return address is the first entry. Then a procedure return is simply

    BR  *callee.staticArea
  

Example

We consider a main program calling a procedure P and then halting. Other actions by Main and P are indicated by subscripted uses of other.

  // Quadruples of Main
  other1
  call P
  other2
  halt
  // Quadruples of P
  other3
  return

Let us arbitrarily assume that the code for Main starts in location 1000 and the code for P starts in location 2000 (there might be other procedures in between). Also assume that each otheri requires 100 bytes (all addresses are in bytes). Finally, we assume that the ARs for Main and P begin at 3000 and 4000 respectively. Then the following machine code results.

  // Code for Main
  1000: Other1
  1100: ST 4000, #1120    // P.staticArea, #here+20
  1112: BR 2000           // Two constants in previous instruction take 8 bytes
  1120: other2
  1220: HALT
        ...
  // Code for P
  2000: other3
  2100: BR *4000
        ...
  // AR for Main
  3000:                   // Return address stored here (not used)
  3004:                   // Local data for Main starts here
        ...
  // AR for P
  4000:                   // Return address stored here
  4004:                   // Local data for P starts here

8.3.2: Stack Allocation

We now need to access the ARs from the stack. The key distinction is that the location of the current AR is not known at compile time. Instead a pointer to the stack must be maintained dynamically.

We dedicate a register, call it SP, for this purpose. In this chapter we let SP point to the bottom of the current AR, that is the entire AR is above the SP. (I do not know why last chapter it was decided to be more convenient to have the stack pointer point to the end of the statically known portion of the activation. However, since the difference between the two is known at compile time it is clear that either can be used.)

The first procedure (or the run-time library code called before any user-written procedure) must initialize SP with
      LD SP, #stackStart
were stackStart is a known-at-compile-time (even -before-) constant.

The caller increments SP (which now points to the beginning of its AR) to point to the beginning of the callee's AR. This requires an increment by the size of the caller's AR, which of course the caller knows.

Is this size a compile-time constant?

Both editions treat it as a constant. The only part that is not known at compile time is the size of the dynamic arrays. Strictly speaking this is not part of the AR, but it must be skipped over since the callee's AR starts after the caller's dynamic arrays.

Perhaps for simplicity we are assuming that there are no dynamic arrays being stored on the stack. If there are arrays, their size must be included in some way.

Caller Calling Callee

The code generated for a parameterless call is

  ADD SP, SP, #caller.ARSize
  ST  *SP, #here+16              // save return address
  BR  callee.codeArea

Callee Returning

The return requires code from both the Caller and Callee. The callee transfers control back to the caller with
      BR *0(SP)
upon return the caller restore the stack pointer with
      SUB SP, SP, caller.ARSize

Example

We again consider a main program calling a procedure P and then halting. Other actions by Main and P are indicated by subscripted uses of `other'.

  // Quadruples of Main
  other[1]
  call P
  other[2]
  halt
  // Quadruples of P
  other[3]
  return

Recall our assumptions that the code for Main starts in location 1000, the code for P starts in location 2000, and each other[i] requires 100 bytes. Let us assume the stack begins at 9000 (and grows to larger addresses) and that the AR for Main is of size 400 (we don't need P.ARSize since P doesn't call any procedures). Then the following machine code results.

  // Code for Main
  1000; LD  SP, 9000
  1008: Other[1]
  1108: ADD SP, SP, #400
  1116: ST  *SP, #1132
  1124: BR, 2000
  1132: SUB SP, SP, #400
  1140: other[2]
  1240: HALT
        ...
  // Code for P
  2000: other[3]
  2100: BR *0(SP)
        ...
  // AR for Main
  9000:                   // Return address stored here (not used)
  9004:                   // Local data for Main starts here
  9396:                   // Last word of the AR is bytes 9396-9399
        ...
  // AR for P
  9400:                   // Return address stored here
  9404:                   // Local data for P starts here

8.3.3: Run-Time Addresses for Names

Basically skipped. A technical fine point about static allocation and (in 1e only) a corresponding point about the display.

Homework: 9.2

8.4: Basic Blocks and Flow Graphs

As we have seen, for may quads it is quite easy to generate a series of machine instructions to achieve the same effect. As we have also seen, the resulting code can be quite inefficient. For one thing the last instruction generated for a quad is often a store of a value that is then loaded right back in the next quad (or one or two quads later).

Another problem is that we don't make much use of the registers. That is translating a single quad needs just one or two registers so we might as well throw out all the other registers on the machine.

Both of the problems are due to the same cause: Our horizon is too limited. We must consider more than one quad at a time. But wild flow of control can make it unclear which quads are dynamically near each other. So we want to consider, at one time, a group of quads within which the dynamic order of execution is tightly controlled. We then also need to understand how execution proceeds from one group to another. Specifically the groups are called basic blocks and the execution order among them is captured by the flow graph.

Definition: A basic block is a maximal collection of consecutive quads such that

  1. Control enters the block only at the first instruction.
  2. Branches (or halts) occur only at the last instruction.

Definition: A flow graph has the basic blocks as vertices and has edges from one block to each possible dynamic successor.

8.4.1: Basic Blocks

Constructing the basic blocks is not hard. Once you find the start of a block, you keep going until you hit a label or jump. But, as usual, to say it correctly takes more words.

Definition: A basic block leader (i.e., first instruction) is any of the following (except for the instruction just past the entire program).

  1. The first instruction of the program.
  2. A target of a (conditional) jump.
  3. The instruction immediately following a (conditional) jump.

Given the leaders, a basic block starts with a leader and proceeds up to but not including the next leader.

Example

The following code produces a 10x10 identity matrix

  for i from 1 to 10 do
    for j from 1 to 10 do
      a[i,j] = 0
    end
  end
  for i from 1 to 10 do
    a[i,i] = 0
  end

The following quads do the same thing

   1)  i = 1
   2)  j = 1
   3)  t1 = 10 * i
   4)  t2 = t1 + j            // element [i,j]
   5)  t3 = 8 * t2            // offset for a[i,j] (8 byte numbers)
   6)  t4 = t3 - 88           // we start at [1,1] not [0,0]
   7)  a[t4] = 0.0
   8)  j = j + 1
   9)  if J <= 10 goto (3)
  10)  i = i + 1
  11)  if i <= 10 goto (2)
  12)  i = 1
  13)  t5 = i - 1
  14)  t6 = 88 * t5
  15)  a[t6] = 1.0
  16)  i = i + 1
  17)  if i <= 10 goto (13)

Which quads are leaders?

1 is a leader by definition. The jumps are 9, 11, and 17. So 10 and 12 are leaders as are the targets 3, 2, and 13.

The leaders are then 1, 2, 3, 10, 12, and 13.

The basic blocks are {1}, {2}, {3,4,5,6,7,8,9}, {10,11}, {12}, and {13,14,15,16,17}.

Here is the code written again with the basic blocks indicated.

   1)  i = 1
  
2) j = 1
3) t1 = 10 * i 4) t2 = t1 + j // element [i,j] 5) t3 = 8 * t2 // offset for a[i,j] (8 byte numbers) 6) t4 = t3 - 88 // we start at [1,1] not [0,0] 7) a[t4] = 0.0 8) j = j + 1 9) if J <= 10 goto (3)
10) i = i + 1 11) if i <= 10 goto (2)
12) i = 1
13) t5 = i - 1 14) t6 = 88 * t5 15) a[t6] = 1.0 16) i = i + 1 17) if i <= 10 goto (13)

We can see that once you execute the leader you are assured of executing the rest of the block in order.

8.4.2: Next Use Information

We want to record the flow of information from instructions that compute a value to those that use the value. One advantage we will achieve is that if we find a value has no subsequent uses, then it is dead and the register holding that value can be used for another value.

Assume that a quad p assigns a value to x (some would call this a def of x).

Definition: Another quad q uses the value computed at p (uses the def) and x is live at q if q has x as an operand and there is a possible execution path from p to q that does not pass any other def of x.

Since the flow of control is trivial inside a basic block, we are able to compute the live/dead status and next use information for at the block leader by a simple backwards scan of the quads (algorithm below).

Note that if x is dead (i.e., not live) on entrance to B the register containing x can be reused in B.

Computing Live/Dead and Next Use Information

Our goal is to determine whether a block uses a value and if so in which statement.

  Initialize all variables in B as being live
  Examine the quads of the block in reverse order.
    Let the quad q compute x and read y and z
    Mark x as dead; mark y and z as live and used at q

When the loop finishes those values that are read before being are marked as live and their first use is noted. The locations x that are set before being read are marked dead meaning that the value of x on entrance is not used.

8.4.3: Flow Graphs

flow graph

The nodes of the flow graph are the basic blocks and there is an edge from P (predecessor) to S (successor) if the last statement of P

  1. is a jump to S or
  2. is not a jump and S immediately follows P.

Two nodes are added: entry and exit. An edge is added from entry to the first basic block.

Edges to the exit are added from any block that could be the last block executed. Specifically edges are added from

  1. the last block if it doesn't end in a jump.
  2. any block that ends in a jump to outside the program.

The flow graph for our example is shown on the right.

8.4.4: Representing Flow Graphs

Note that jump targets are no longer quads but blocks. The reason is that various optimizations within blocks will change the instructions and we would have to change the jump to reflect this.

8.4.5: Loops

Of course most of a program's execution time is within loops so we want to identify these.

Definition: A collection of basic blocks forms a loop L with loop entry E if

  1. no block other than E has a predecessor outside L.
  2. all blocks in L have a path to E completely inside L.

The flow graph on the right has three loops.

  1. {B3}, i.e., B3 by itself.
  2. {B6}.
  3. {B2, B3, B4}

Homework: Consider the following program for matrix multiplication.

  for (i=0; i<10; i++)
      for (j=0; j<10; j++)
          c[i][j] = 0;
  for (i=0; i<10; i++)
      for (j=0; j<10; j++)
          for (k=0; k<10; k++)
              c[i][j] = c[i][j] + a[i][k] * b[k][j];
  1. Produce quads of the form we have been using. Assume each element requires 8 bytes.
  2. What are the basic block of your program?
  3. Construct the flow graph.
  4. Identify the loops in your flow graph.

================ Start Lecture #14 ================

A Word or Two About Global Flow Analysis (unofficial)

We are not covering global flow analysis; it is a key component of optimization and would be a natural topic in a follow-on course. Nonetheless there is something we can say just by examining the flow graphs we have constructed. For this discussion I am ignoring tricky and important issues concerning arrays and pointer references (specifically, disambiguation). You may wish to assume that the program contains no arrays or pointers for these comments.

We have seen that a simple backwards scan of the statements in a basic block enables us to determine the variables that are live-on-entry and those that are dead-on-entry. Those variables that do not occur in the block are in neither category; perhaps we should call them ignored by the block.

We shall see below that it would be lovely to know which variables are live/dead-on-exit. This means which variables hold values at the end of the block that will / will not be used. To determine the status of v on exit of a block B, we need to trace all possible execution paths beginning at the end of B. If all these paths reach a block where v is dead-on-entry before they reach a block where v is live-on-entry, then v is dead on exit for block B.

8.5: Optimization of Basic Blocks

8.5.1: The DAG Representation of Basic Blocks

The goal is to obtain a visual picture of how information flows through the block. The leaves will show the values entering the block and as we proceed up the DAG we encounter uses of these values defs (and redefs) of values and uses of the new values.

Formally, this is defined as follows.

  1. Create a leaf for the initial value of each variable appearing in the block. (We do not know what that the value is, not even if the variable has ever been given a value).
  2. Create a node N for each statement s in the block.
    1. Label N with the operator of s. This label is drawn inside the node.
    2. Attach to N those variables for which N is the last def in the block. These additional labels are drawn along side of N.
    3. Draw edges from N to each statement that is the last def of an operand used by N.
  3. Designate as output nodes those N whose values are live on exit, an officially-mysterious term meaning values possibly used in another block. (Determining the live on exit values requires global, i.e., inter-block, flow analysis.)

As we shall see in the next few sections various basic-block optimizations are facilitated by using the DAG. DAG 8.12

8.5.2: Finding Local Common Subexpressions

As we create nodes for each statement, proceeding in the static order of the statements, we might notice that a new node is just like one already in the DAG in which case we don't need a new node and can use the old node to compute the new value in addition to the one it already was computing.

Specifically, we do not construct a new node if an existing node has the same children in the same order and is labeled with the same operation.

Consider computing the DAG for the following block of code.

    a = b + c
    c = a + x
    d = b + c
    b = a + x
  

The DAG construction is explain as follows (the movie on the right accompanies the explanation).

  1. First we construct leaves with the initial values.
  2. Next we process a = b + c. This produces a node labeled + with a attached and having b0 and c0 as children.
  3. Next we process c = a + x.
  4. Next we process d = b + c. Although we have already computed b + c in the first statement, the c's are not the same, so we produce a new node.
  5. Then we process b = a + x. Since we have already computed a + x in statement 2, we do not produce a new node, but instead attach b to the old node.
  6. Finally, we tidy up and erase the unused initial values.

You might think that with only three computation nodes in the DAG, the block could be reduced to three statements (dropping the computation of b). However, this is wrong. Only if b is dead on exit can we omit the computation of b. We can, however, replace the last statement with the simpler
      b = c.

Sometimes a combination of techniques finds improvements that no single technique would find. For example if a-b is computed, then both a and b are incremented by one, and then a-b is computed again, it will not be recognized as a common subexpression even though the value has not changed. However, when combined with various algebraic transformations, the common value can be recognized.

8.5.3: Dead Code Elimination

DAG 8.13

Assume we are told (by global flow analysis) that certain values are dead on exit. We examine each root (node with no ancestor) and delete any that have no live variables attached. This process is repeated since new roots may have appeared.

For example, if we are told, for the picture on the right, that only a and b are live, then the root d can be removed since d is dead. Then the rightmost node becomes a root, which also can be removed (since c is dead).

8.5.4: The Use of Algebraic Identities

Some of these are quite clear. We can of course replace x+0 or 0+x by simply x. Similar considerations apply to 1*x, x*1, x-0, and x/1.

Another class of simplifications is strength reduction, where we replace one operation by a cheaper one. A simple example is replacing 2*x by x+x on architectures where addition is cheaper than multiplication.

A more sophisticated strength reduction is applied by compilers that recognize induction variables (loop indices). Inside a
    for i from 1 to N
loop, the expression 4*i can be strength reduced to j=j+4 and 2^i can be strength reduced to j=2*j (with suitable initializations of j just before the loop).

Other uses of algebraic identities are possible; many require a careful reading of the language reference manual to ensure their legality. For example, even though it might be advantageous to convert
    ((a + b) * f(x)) * a
to
    ((a + b) * a) * f(x)
it is illegal in Fortran since the programmer's use of parentheses to specify the order of operations can not be violated.

Does

    a = b + c
    x = y + c + b + r
  
contain a common subexpression of b+c that need be evaluated only once?
The answer depends on whether the language permits the use of the associative and commutative law for addition. (Note that the associative law is invalid for floating point numbers.)

8.5.5: Representation of Array References

Arrays are tricky. Question: Does

    x = a[i]
    a[j] = 3
    z = a[i]
  
contain a common subexpression of a[i] that need be evaluated only once?
The answer depends on whether i=j. Without some form of disambiguation, we can not be assured that the values of i and j are distinct. Thus we must support the worst case condition that i=j and hence the two evaluations of a[i] must each be performed.

A statement of the form x = a[i] generates a node labeled with the operator =[] and the variable x, and having children a0, the initial value of a, and the value of i.

DAG 8.14

A statement of the form a[j] = y generates a node labeled with operator []= and three children a0. j, and y, but with no variable as label. The new feature is that this node kills all existing nodes depending on a0. A killed node can not received any future labels so cannot becomew a common subexpression.

Returning to our example

    x = a[i]
    a[j] = 3
    z = a[i]
  

We obtain the top figure to the right.

DAG 8.15

Sometimes it is not children but grandchildren (or other descendant) that are arrays. For example we might have

    b = a + 8    // b[i] is 8 bytes past a[i]
    x = b[i]
    b[j] = y
  
Again we need to have the third statement kill the second node even though it is caused by a grandchild. This is shown in the bottom figure.

8.5.6: Pointer Assignment and Procedure Calls

Pointers are even trickier than arrays. Together they have spawned a mini-industry in disambiguation, i.e., when can we tell whether two array or pointer references refer to the same or different locations. A trivial case of disambiguation occurs with.

    p = &x
    *p = y
  
In this case we know precisely the value of p so the second statement kills only nodes with x attached.

With no disambiguation information, we must assume that a pointer can refer to any location. Consider

    x = *p
    *q = y
  

We must treat the first statement as a use of every variable; pictorially the =* operator takes all current nodes with identifiers as arguments. This impacts dead code elimination.

We must treat the second statement as writing every variable. That is all existing nodes are killed, which impacts common subexpression elimination.

In our basic-block level approach, a procedure call has properties similar to a pointer reference: For all x in the scope of P, we must treat a call of P as using all nodes with x attached and also kills those same nodes.

8.5.7: Reassembling Basic Blocks From DAGs

Now that we have improved the DAG for a basic block, we need to regenerate the quads. That is, we need to obtain the sequence of quads corresponding to the new DAG.

We need to construct a quad for every node that has a variable attached. If there are several variables attached we chose a live-on-exit variable, assuming we have done the necessary global flow analysis to determine such variables).

If there are several live-on-exit variables we need to compute one and make a copy so that we have both. An optimization pass may eliminate the copy if it is able to assure that one such variable may be used whenever the other is referenced.

Example

Recall the example from our movie

    a = b + c
    c = a + x
    d = b + c
    b = a + x
  

If b is dead on exit, the first three instructions suffice. If not we produce instead

    a = b + c
    c = a + x
    d = b + c
    b = c
  
which is still an improvement as the copy instruction is less expensive than the addition on most architectures.

If global analysis shows that, whenever this definition of b is used, c contains the same value, we can eliminate the copy and use c in place of b.

Order of Generated Instructions

Note that of the following 5, rules 2 are due to arrays, and 2 due to pointers.

  1. The DAG order must be respected (defs before uses).
  2. Assignment to an array must follow all assignments to or uses of the same array that preceded it in the original block (no reordering of array assignments).
  3. Uses of an array must follow all (preceding according to the original block) assignments to it; so the only transformation possible is reordering uses.
  4. All variable references must follow all (preceding ...) procedure calls or assignment through a pointer.
  5. A procedure call or assignment through a pointer must follow all (preceding ...) variable references.

Homework: 9.14,
9.15 (just simplify the 3-address code of 9.14 using the two cases given in 9.15), and
9.17 (just construct the DAG for the given basic block in the two cases given).

8.6: A Simple Code Generator

A big issue is proper use of the registers, which are often in short supply, and which are used/required for several purposes.

  1. Some operands must be in registers.
  2. Holding temporaries thereby avoiding expensive memory ops.
  3. Holding inter-basic-block values (loop index).
  4. Storage management (e.g., stack pointer).

For this section we assume a RISC architecture. Specifically, we assume only loads and stores touch memory; that is, the instruction set consists of

    LD  reg, mem
    ST  mem, reg
    OP  reg, reg, reg
  
where there is one OP for each operation type used in the three address code.

The 1e uses CISC like instructions (2 operands). Perhaps 2e switched to RISC in part due to the success of the ROPs in the Pentium Pro.

A major simplification is we assume that, for each three address operation, there is precisely one machine instruction that accomplishes the task. This eliminates the question of instruction selection.

We do, however, consider register usage. Although we have not done global flow analysis (part of optimization), we will point out places where live-on-exit information would help us make better use of the available registers.

Recall that the mem operand in the load LD and store ST instructions can use any of the previously discussed addressing modes. addressing modes

Addressing Mode Usage

Remember that in 3-address instructions, the variables written are addresses, i.e., they represent l-values.

Let us assume a is 500 and b is 700, i.e., a and b refer to locations 500 and 700 respectively. Assume further that location 100 contains 666, location 500 contains 100, location 700 contains 900, and location 900 contains 123. This initial state is shown in the upper left picture.

In the four other pictures the contents of the pink location has been changed to the contents of the light green location. These correspond to the three-address assignment statements shown below each picture. The machine instructions indicated below implement each of these assignment statements.

  a = b
      LD  R1, b
      ST  a, R1

  a = *b
      LD  R1, b
      LD  R1, 0(R1)
      ST  a, R1

  *a = b
      LD  R1, b
      LD  R2, a
      ST  0(R2), R1

  *a = *b
      LD  R1, b
      LD  R1, 0(R1)
      LD  R2, a
      ST  0(R2), R1

8.6.1: Register and Address Descriptors

These are the primary data structures used by the code generator. They keep track of what values are in each register as well as where a given value resides.

The register descriptor could be omitted since you can compute it from the address descriptors.

8.6.2: The Code-Generation Algorithm

There are basically three parts to (this simple algorithm for) code generation.

  1. Choosing registers
  2. Generating instructions
  3. Managing descriptors

We will isolate register allocation in a function getReg(Instruction), which is presented later. First presented is the algorithm to generate instructions. This algorithm uses getReg() and the descriptors. Then we learn how to manage the descriptors and finally we study getReg() itself.

Machine Instructions for Operations

Given a quad OP x, y, z (i.e., x = y OP z), proceed as follows.

  1. Call getReg(OP x, y, z) to get Rx, Ry, and Rz, the registers to be used for x, y, and z respectively.

    Note that getReg merely selects the registers, it does not guarantee that the desired values are present in these registers.

  2. Check the register descriptor for Ry. If y is not present in Ry, check the address descriptor for y and issue
          LD Ry, y

    The 2e uses y' (not y) as source of the load, where y' is some location containing y (1e suggests this as well). I don't see how the value of y can appear in any memory location other than y. Please check me on this.

    One might worry that either

    1. The value is located nowhere or
    2. The value is located in a register different from Ry.

    It would be a serious bug in the algorithm if the first were true, and I am confident it is not. The second might be a possible design, but when we study getReg(), we will see that if the value of y is in some register, then the chosen Ry will contain that value.

  3. Similar treatment for Rz.

  4. Generate the instruction
          OP Rx, Ry, Rz

Machine Instructions for Copy Statements

When processing
      x = y
steps 1 and 2 are the same as above (getReg() will set Rx=Ry). Step 3 is vacuous and step 4 is omitted. This says that if y was already in a register before the copy instruction, no code is generated at this point. Since the value of y is not in its memory location, we may need to store this value back into y at block exit.

Ending the Basic Block

You probably noticed that we have not yet generated any store instructions; They occur here (and during spill code in getReg()). We need to ensure that all variables needed by (dynamically) subsequent blocks (i.e., those live-on-exit) have their current values in their memory locations.

  1. Temporaries are never live beyond a basic block so can be ignored.

  2. Variables dead on exit (thank you global flow for determining such variables) are also ignored.

  3. All live on exit variables (for us all non-temporaries) need to be in their memory location on exit from the block.

    Check the address descriptor for each live on exit variable. If its own memory location is not listed, generate
          ST x, R
    where R is a register listed in the address descriptor

Managing Register and Address Descriptors

This is fairly clear. We just have to think through what happens when we do a load, a store, an OP, or a copy. For R a register, let Desc(R) be its register descriptor. For x a program variable, let Desc(x) be its address descriptor.

  1. Load: LD R, x
  2. Store: ST x, R
  3. Operation: OP Rx, Ry, Rz implementing the quad OP x, y, z
  4. Copy: For x = y after processing the load (if needed)

Example

Since we haven't specified getReg() yet, we will assume there are an unlimited number of registers so we do not need to generate any spill code (saving the register's value in memory). One of getReg()'s jobs is to generate spill code when a register needs to be used for another purpose and the current value is not presently in memory.

Despite having ample registers and thus not generating spill code, we will not be wasteful of registers.

This example is from the book. I give another example after presenting getReg(), that I believe justifies my claim that the book is missing an action, as indicated above.

Assume a, b, c, and d are program variables and t, u, v are compiler generated temporaries (I would call these t$1, t$2, and t$3). The intermediate language program is on the left with the generated code for each quad shown. To the right is shown the contents of all the descriptors. The code generation is explained below the diagram.

dia 8.16



       t = a - b
           LD  R1, a
           LD  R2, b
           SUB R2, R1, R2


       u = a - c
           LD  r3, c
           SUB R1, R1, R3

       v = t + u
           ADD R3, R2, R1


       a = d
           LD  R2, d


       d = v + u
           ADD R1, R3, R1


       exit
           ST  a, R2
           ST  d, R1

What follows describes the choices made. Confirm that the values in the descriptors matches the explanations.

  1. For the first quad, we need all three instructions since nothing is register resident on block entry. Since b is not used again, we can reuse its register. (Note that the current value of b is in its memory location.)
  2. We do not load a again since its value is R1, which we can reuse for u since a is not used below.
  3. We again reuse a register for the result; this time because c is not used again.
  4. The copy instruction required a load since d was not in a register. As the descriptor shows, a was assigned to the same register, but no machine instruction was required.
  5. The last instruction uses values already in registers. We can reuse R1 since u is a temporary.
  6. At block exit, lacking global flow analysis, we must assume all program variables are live and hence must store back to memory any values located only in registers.

8.6.3: Design of the Function getReg

Consider
      x = y OP z
Picking registers for y and z are the same; we just do y. Choosing a register for x is a little different.

A copy instruction
      x = y
is easier.

Choosing Ry

Similar to demand paging, where the goal is to produce an available frame, our objective here is to produce an available register we can use for Ry. We apply the following steps in order until one succeeds. (Step 2 is a special case of step 3.)

  1. If Desc(y) contains a register, use of these for Ry.
  2. If Desc(R) is empty for some registers, pick one of these.
  3. Pick a register for which the cleaning procedure generates a minimal number of store instructions. To clean an in-use register R do the following for each v in Desc(R).
    1. If Desc(v) includes something besides R, no store is needed for v.
    2. If v is x and x is not z, no store is needed since x is being overwritten.
    3. No store is needed if there is no further use of v prior to a redefinition. This is easy to check for further uses within the block. If v is live on exit (e.g., we have no global flow analysis), we need a redefinition later in this block.
    4. Otherwise a spill ST v, R is generated.

Choosing Rz and Rx, and Processing x = y

As stated above choosing Rz is the same as choosing Ry.

Choosing Rx has the following differences.

  1. Since Rx will be written it is not enough for Desc(x) to contain a register R as in 1. above; instead, Desc(R) must contain only x.
  2. If there is no further use of y prior to a redefinition (as described above for v) and if Ry contains only y (or will do so after it is loaded), then Ry can be used for Rx. Similarly, Rz might be usable for Rx.

getReg(x=y) chooses Ry as above and chooses Rx=Ry.

Example

                            R1  R2  R3    a    b    c    d    e
			                  a    b    c    d    e

  a = b + c
      LD  R1, b
      LD  R2, c
      ADD R3, R1, R2
                            R1  R2  R3    a    b    c    d    e
			    b   c   a     R3  b,R1 c,R2  d    e

  d = a + e
      LD  R1, e
      ADD R2, R3, R1
                            R1  R2  R3    a    b    c    d    e
                     2e →   e   d   a     R3  b,R1  c    R2  e,R1
                     me →   e   d   a     R3   b    c    R2  e,R1

We needed registers for d and e; none were free. getReg() first chose R2 for d since R2's current contents, the value of c, was also located in memory. getReg() then chose R1 for e for the same reason.

Using the 2e algorithm, b might appear to be in R1 (depends if you look in the address or register descriptors).

  a = e + d
      ADD R3, R1, R2
                            Descriptors unchanged

  e = a + b
      ADD R1, R3, R1   ← possible wrong answer from 2e
                            R1  R2  R3    a    b    c    d    e
                            e   d    a    R3  b,R1  c    R2   R1

      LD  R1, b
      ADD R1, R3, R1
                            R1  R2  R3    a    b    c    d    e
                            e   d    a    R3   b    c    R2   R1

The 2e might think R1 has b (address descriptor) and also conclude R1 has only e (register descriptor) so might generate the erroneous code shown.

Really b is not in a register so must be loaded. R3 has the value of a so was already chosen for a. R2 or R1 could be chosen. If R2 was chosen, we would need to spill d (we must assume live-on-exit, since we have no global flow analysis). We choose R1 since no spill is needed: the value of e (the current occupant of R1) is also in its memory location.

  exit
      ST  a, R3
      ST  d, R2
      ST  e, R1

8.7: Peephole Optimization

Skipped.

8.8: Register Allocation and Assignment

Skipped.

8.9: Instruction Selection by Tree Rewriting (unofficial)

What if a given quad needs several OPs and we have choices?

We would like to be able to describe the machine OPs in a way that enables us to find a sequence of OPs (and LDs and STs) to do the job.

The idea is that you express the quad as a tree and express each OP as a (sub-)tree simplification, i.e. the op replaces a subtree by a simpler subtree. In fact the simpler subtree is just a single node. tree 8.9

The diagram on the right represents x[i] = y[a] + 9, where x and y are on the stack and a is in the static area. M's are values in memory; C's are constants; and R's are registers. The weird ind (presumably short for indirect) treats its argument as a memory location.

Compare this to grammars: A production replaces the RHS by the LHS. We consider context free grammars where the LHS is a single nonterminal.

For example, a LD replaces a Memory node with a Register node.

Another example is that ADD Ri, Ri, Rj replaces a subtree consisting of a + with both children registers (i and j) with a Register node (i).

As you do the pattern matching and reductions (apply the productions), you emit the corresponding code (semantic actions). So to support a new processor, you need to supply the tree transformations corresponding to every instruction in the instruction set.

8.10: Optimal Code Generation for Expressions (unofficial)

This is quite cute.

We assume all operators are binary and label the instruction tree with something like the height. This gives the minimum number of registers needed so that no spill code is required. A few details follow.

8.10.1: Ershov Numbers

  1. Draw the expression tree.
  2. Label the leaves with 1.
  3. Label interior nodes with L:
    1. If the children have the same label x, L=x+1. This looks like height.
    2. If the children have different labels, x and y, L=max(x,y).

8.10.2: Generating Code From Labeled Expression Trees

  1. Recursive algorithm starting at the root. Each node puts its answer in the highest number register it is assigned. The idea is that a node uses (mostly) the same registers as its sibling.
    1. If the labels on the children are equal to L, the parent's label is L+1.
      1. Give one child L regs answer appears in top reg.
      2. Give other child L regs, but one higher, answer again appears in top reg.
      3. Parent uses a two address OP to compute answer in the same reg used by second child, which is the top reg assigned to the parent.
    2. If the labels on the children are M<L, the parent is labeled L.
      1. Give bigger child L regs.
      2. Give other child M regs ending one below bigger child.
      3. Parent uses 2-addr OP computing answer in L
    3. If at a leaf (operand), load it into assigned reg.

Can see this is optimal (assuming you have enough registers).

  1. Loads each operand only once.
  2. Performs each operation only once.
  3. Does no stores.
  4. Minimal number of registers having the above three properties.
    1. Show need L registers to produce a result with label L.
    2. Must compute one side and no use the register containing its answer before finishing the other side.
    3. Apply this argument recursively.

8.10.3: Evaluating Expressions with an Insufficient Supply of Registers

Rough idea is to apply the above recursive algorithm, but at each recursive step, if the number of regs is not enough, store the result of the first child computed before starting the second.

8.11: Dynamic Programming Code-Generation

Skipped