C program compiler presentation

Rigvendra Kumar Vardhan
M.TECH ECE
Pondicherry University
1

 Interpreters (direct execution)
 Assemblers
 Preprocessors
 Text formatters (non-WYSIWYG)
 Analysis tools
CS 540 Spring 2013 GMU 2

 Interpreter
A program that reads a source program
and produces the results of executing
that program
 Compiler
A program that translates a program
from one language (the source) to
another (the target)
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 Correctness
 Speed (runtime and compile time)
Degrees of optimization
Multiple passes
 Space
 Feedback to user
 Debugging

 Interpreter
Execution engine
Program execution interleaved with
analysis
running = true;
while (running) {
analyze next statement;
execute that statement;
}
May involve repeated analysis of some
statements (loops, functions)
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 Read and analyze entire program
 Translate to semantically equivalent
program in another language
Presumably easier to execute or more
efficient
Should “improve” the program in some
fashion
 Offline process
Tradeoff: compile time overhead
(preprocessing step) vs execution
performance
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 Compilers
FORTRAN, C, C++, Java, COBOL, etc.
etc.
Strong need for optimization, etc.
 Interpreters
PERL, Python, awk, sed, sh, csh,
postscript printer, Java VM
Effective if interpreter overhead is low
relative to execution cost of language
statements
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8
Source code
Compiler
Assembly code
Assembler
Object code
(machine code)
Fully-resolved object
code (machine code)
Linker
Loader
Executable image

 Series of program representations
 Intermediate representations optimized
for program manipulations of various
kinds (checking, optimization)
 Become more machine-specific, less
language-specific as translation proceeds
9

 First approximation
Front end: analysis
 Read source program and understand its structure
and meaning
Back end: synthesis
 Generate equivalent target language program
Source Front End Back End Target
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 Must recognize legal programs (&
complain about illegal ones)
 Must generate correct code
 Must manage storage of all variables
 Must agree with OS & linker on target
format
11

 Need some sort of Intermediate
Representation (IR)
 Front end maps source into IR
 Back end maps IR to target machine code
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Scanner
(lexical
analysis)
Parser
(syntax
analysis)
Code
Optimizer
Semantic
Analysis
(IC generator)
Code
Generator
Symbol
Table
Source
language
tokens Syntactic
structure
Intermediate
Language
Target
language
Intermediate
Language

 Lexical Analysis
 Syntax Analysis
 Semantic Analysis
 Runtime environments
 Code Generation
 Code Optimization

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Source code
(character stream)
Lexical analysis
Parsing
Token stream
Abstract syntax tree
Intermediate Code Generation
Intermediate code
Optimization
Code generation
Intermediate code
Assembly code
Front end
(machine-independent)
Back end
(machine-dependent)

Scanner Parser source tokens IR
 Split into two parts
Scanner: Responsible for converting
character stream to token stream
 Also strips out white space, comments
Parser: Reads token stream; generates IR
 Both of these can be generated
automatically
Source language specified by a formal
grammar
Tools read the grammar and generate
scanner & parser (either table-driven or hard
coded)
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 Token stream: Each significant lexical
chunk of the program is represented by a
token
Operators & Punctuation: {}[]!+-=*;: …
Keywords: if while return goto
Identifiers: id & actual name
Constants: kind & value; int, floating-point
character, string, …
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 Input text
// this statement does very little
if (x >= y) y = 42;
 Token Stream
IF LPAREN ID(x) GEQ ID(y)
RPAREN ID(y) BECOMES INT(42) SCOLON
Note: tokens are atomic items, not character
strings
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 Many different forms
(Engineering tradeoffs)
 Common output from a parser is an
abstract syntax tree
Essential meaning of the program
without the syntactic noise
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 Token Stream Input  Abstract Syntax Tree
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IF LPAREN ID(x)
GEQ ID(y) RPAREN
ID(y) BECOMES
INT(42) SCOLON
ifStmt
>=
ID(x) ID(y)
assign
ID(y) INT(42)

 During or (more common) after parsing
Type checking
Check for language requirements like
“declare before use”, type compatibility
Preliminary resource allocation
Collect other information needed by
back end analysis and code generation
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 Responsibilities
Translate IR into target machine code
Should produce fast, compact code
Should use machine resources
effectively
Registers
Instructions
Memory hierarchy
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 Typically split into two major parts with
sub phases
“Optimization” – code improvements
May well translate parser IR into
another IR
Code generation
Instruction selection & scheduling
Register allocation
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 Input
if (x >= y)
y = 42;
 Output
mov eax,[ebp+16]
cmp eax,[ebp-8]
jl L17
mov [ebp-8],42
L17:
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lda $30,-32($30)
stq $26,0($30)
stq $15,8($30)
bis $30,$30,$15
bis $16,$16,$1
stl $1,16($15)
lds $f1,16($15)
sts $f1,24($15)
ldl $5,24($15)
bis $5,$5,$2
s4addq $2,0,$3
ldl $4,16($15)
mull $4,$3,$2
ldl $3,16($15)
addq $3,1,$4
mull $2,$4,$2
ldl $3,16($15)
addq $3,1,$4
mull $2,$4,$2
stl $2,20($15)
ldl $0,20($15)
br $31,$33
$33:
bis $15,$15,$30
ldq $26,0($30)
ldq $15,8($30)
addq $30,32,$30
ret $31,($26),1
Optimized Code
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s4addq $16,0,$0
mull $16,$0,$0
addq $16,1,$16
mull $0,$16,$0
mull $0,$16,$0
ret $31,($26),1
Unoptimized Code

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Source code
(character stream)
Lexical analysis
Parsing
Token stream
Abstract syntax tree
(AST)
Semantic Analysis
if (b == 0) a = b;
if ( b == 0 ) a = b ;
if
==
b 0
=
a b
if
==
int b int 0
=
int a
lvalue
int b
boolean
Decorated AST
;
int ;

Intermediate Code Generation
Optimization
Code generation
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if
boolean ==
int ;
int b int 0
=
int a
lvalue
int b
CJUMP ==
MEM
+
fp 8
CONST MOVE
0 MEM MEM
fp 4 fp 8
NOP
+ +
CJUMP ==
CONST MOVE
CX NOP
0 DX CX
CMP CX, 0
CMOVZ DX,CX

 Compiler techniques are everywhere
Parsing (little languages, interpreters)
Database engines
AI: domain-specific languages
Text processing
Tex/LaTex -> dvi -> Postscript -> pdf
Hardware: VHDL; model-checking tools
Mathematics (Mathematica, Matlab)
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 Fascinating blend of theory and
engineering
Direct applications of theory to practice
Parsing, scanning, static analysis
Some very difficult problems (NP-hard or
worse)
Resource allocation, “optimization”,
etc.
Need to come up with good-enough
solutions
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 Ideas from many parts of CSE
AI: Greedy algorithms, heuristic search
Algorithms: graph algorithms, dynamic
programming, approximation
algorithms
Theory: Grammars DFAs and PDAs,
pattern matching, fixed-point
algorithms
Systems: Allocation & naming,
synchronization, locality
Architecture: pipelines & hierarchy
management, instruction set use
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 program ::= statement | program
statement
 statement ::= assignStmt | ifStmt
 assignStmt ::= id = expr ;
 ifStmt ::= if ( expr ) stmt
 expr ::= id | int | expr + expr
 Id ::= a | b | c | i | j | k | n | x | y | z
 int ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
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 There are several syntax notations for
productions in common use; all mean the
same thing
ifStmt ::= if ( expr ) stmt
ifStmt if ( expr ) stmt
<ifStmt> ::= if ( <expr> ) <stmt>
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program ::= statement | program statement
statement ::= assignStmt | ifStmt
assignStmt ::= id = expr ;
ifStmt ::= if ( expr ) stmt
expr ::= id | int | expr + expr
id ::= a | b | c | i | j | k | n | x | y | z
int ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
a = 1 ;
if ( a + 1 )
b = 2 ;
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program
program
ID(a) expr
stmt
stmt
assign
int (1)
ifStmt
expr stmt
expr expr
assign
ID(a) int (1) ID(b) expr
int (2)

C program compiler presentation

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