2_SimpleOnePassCompiler

8/7/2019 2_SimpleOnePassCompiler

1/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 1/66

2. A SIMPLE ONE-PASS COMPILERInfix expression -> Postfix expression

Yoon-Joong KimComputer Engineering Department, Hanbat National University


2/66


2.1 Overview :

Language: Appearance(Vocabulay,Syntax), Semantics

2.2 Syntax definition

Alphabet, string(sentence), language, BNF, Context Free Grammar, Parse Tree, Ambiguity,

Associativity and precedence of operator, Syntax of assignment statement.

2.3 Syntax-directed translation :

SDD, SDTS

2.4 Parsing :Top-Down Parsing, Predictive Parsing(Recursive Decent Parsing), Left factoring, Elimination of left-recursion

2.5 A translator for simple expression : Design and implementation

2.6 Lexical analysis

Design and implementation of as simple lexical analyzer

2.7 Incorporating a symbol table

2.8 A abstract stack machine

Stack machine operation, Lvalue, Rvalue, Stack manipulation, Translation of expression,

Translation of statements. Emitting a translation


3/66


2.9 Putting the techniques together

Design and implementation of a simple one pass translator, infix expression to postfix expression


4/66


This chapter

is an introduction to the material in Chapter 3 through 8. concentrates on the front end of compiler, that is, on lexical analysis, parsing, and

intermediate code generation.

2.1 OVERVIEWLanguage Definition

- Appearance of programming language :

Vocabulary : Regular expressionSyntax : Backus-Naur Form(BNF) or Context Free Form(CFG)

- Semantics : Informal language or some examples

l Fig 2.1. Structure of our compiler front end


5/66


2.2 SYNTAX DEFINITION

lTo specify the syntax of a language : CFG and BNF Example : if-else statement in C has the form of

statement if ( expression ) statement else statementAn alphabet of a language is a set of symbols.

Examples : {0,1} for a binary number system(language)={0,1,100,101,...}

{a,b,c} for language={a,b,c, ac,abcc..}

{if ,(,),else ...} for a if statements={if(a==1)goto10, if--}A string over an alphabet

is a sequence of zero or more symbols from the alphabet.

Examples : 0,1,10,00,11,111,0202 ... strings for a alphabet {0,1}

Null string is a string which does not have any symbol of alphabet.

Language

Is a subset of all the strings over a given alphabet. Alphabets Ai Languages Li for Ai

A0={0,1} L0={0,1,100,101,...}


6/66


A1={a,b ,c} L1={a,b,c, ac, abcc..}

A2={all of C tokens} L2= {all sentences of C program }

Example 2.1. Grammar for expressions consisting of digits and plus and minus

signs.

Language of expressions L={9-5+2, 3-1, ...}

The productions of grammar for this language L are:

list list + digit

list list - digit

list digit

digit 0|1|2|3|4|5|6|7|8|9

list, digit : Grammar variables, Grammar symbols

{0,1,2,3,4,5,6,7,8,9,-,+} : Alphabet for language L, Tokens, Terminal symbolsConvention specifying grammar

Terminal symbols : bold face string i f , num, id Nonterminal symbol, grammar symbol : italicized names, list, digit ,A,B


7/66


Grammar G=(N,T,P,S)

N : a set of nonterminal symbols

T : a set of terminal symbols, tokens

P : a set of production rules

S : a start symbol, SN

Grammar G for a language L={9-5+2, 3-1, ...}

G=(N,T,P,S)

N={list,digit}T={0,1,2,3,4,5,6,7,8,9,-,+} : is a kind of alphabet for L

P : list -> list + digit list -> list - digit list -> digit

digit -> 0 |1 |2 |3 |4 |5 |6 |7 |8 |9S= list


8/66


Some definitions for a language L and its grammar G

Derivation :A sequence of replacements S 1 2n is a derivation of n.

Example, A derivation 1+9 from the grammar G

left most derivation list list + digit digit + digit 1 + digit 1 + 9

right most derivation list list + digit list + 9 digit + 9 1 + 9

Language of grammar L(G)L(G) is a set of sentences(strings) that can be generated from the grammar G=(N,T,P,S).

L(G)={x| S *

x} where x a sequence of terminal symbols

Example: Consider a grammar G=(N,T,P,S):

N={S} T={a,b}

S=S P ={S aSb | }

Is aabb a sentence of L(g)? (derivation of string aabb)SaSbaaSbbaabbaabb(or S* aabb) so, aabbL(G)

there is no derivation for aa, so aaL(G) note L(G)={anbn| n0} where anbn meas n as followed by n bs.


9/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 9/66Parse TreeA derivation can be conveniently represented by a derivation tree( parse tree).

The root is labeled by the start symbol.

Each leaf is labeled by a token or .

Each interior nodes are labeled by a nonterminal symbol.

When a production Ax 1 xn is derived, nodes labeled by x 1 xn are made as children

nodes of node labeled by A.

root : the start symbol internal nodes : nonterminal leaf nodes : terminal


10/66


Example G:

list -> list + digit | list - digit | digit

digit -> 0|1|2|3|4|5|6|7|8|9

left most derivation for 9-5+2,list list+digit list-digit+digit digit-digit+digit 9-digit+digit

9-5+digit 9-5+2 right most derivation for 9-5+2,

list list+digit list+2 list-digit+2 list-5+2 digit-5+2 9-5+2

parse tree for 9-5+2


11/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 11/66AmbiguityA grammar is said to be ambiguous if the grammar has more than one parsetree for a given string of tokens.

Example 2.5. Suppose a grammar G that can not distinguish between lists anddigits as in Example 2.1.

G : string string + string | string - string |0|1|2|3|4|5|6|7|8|9

1-5+2 has 2 parse trees => Grammar G is ambiguous.


12/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 12/66Associativity of operatorA operator is said to be left associative if an operand with operators on both sides of it is taken bythe operator to its left.

eg) 9+5+2(9+5)+2, a=b=ca=(b=c)

Left Associative Grammar :

list list + digit | list - digit

digit 0|1||9

Right Associative Grammar :

right letter = right | letter letter a|b||z

left associative operators :+ , - , * , /

right associative operators := , **


13/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 13/66Precedence of operatorsWe say that a operator(*) has higher precedence than other operator(+) if the operator(*) takesoperands before other operator(+) does.

ex. 9+5*29+(5*2), 9*5+2(9*5)+2

Syntax of full expressions

operator associative precedence+ , - left 1 low

* , / left 2 heigh

left a. left a. precedence/depth

expr expr + term | expr - term | term 1

term term * factor | term / factor | factor 2

factor digit | ( expr ) 3digit 0 | 1 | | 9 4


14/66


15/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 15/662.3 SYNTAX-DIRECTED TRANSLATION(SDT)A formalism for specifying translations for programming language constructs.

( attributes of a construct: type, string, location, etc)

Syntax directed definition(SDD) for the translation of constructs Syntax directed translation scheme(SDTS) for specifying translation

Postfix notation for an (infix) expression E If E is a variable or constant, then the postfix nation for E is E itself ( E.tE ).

if E is an expression of the form E1 op E2 where op is a binary operator

E1' is the postfix of E1, E2' is the postfix of E2 then E1' E2' op is the postfix for E1 op E2

if E is (E1), and E 1' is a postfix

then E1' is the postfix for E

eg) 9 - 5 + 2 9 5 - 2 +

9 - (5 + 2) 9 5 2 + -


16/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 16/66Syntax-Directed Definition(SDD) for translationSDD is a set of semantic rules predefined for each productions respectively fortranslation.

A translation is an input-output mapping procedure for translation of an input X,

construct a parse tree for X.

synthesize attributes in order of depth-first traversal over the parse tree.

Suppose a node n in parse tree is labeled by X and X.a denotes the value of attribute aof X at that node.

compute X's attributes X.a using the semantic rules associated with X.

Depth-First Traversalsprocedure visit(n: node)

beginfo r each child m of n, from left to right do visit(m);

evaluate semantic rules at node nend


17/66


An example of translating input X=9-5+2 to postfix expression (synthesized attributes for input X=9-5+2)

1. produce parse tree for the input X=9-5+22. evaluate attribute of node according to the dept-first order based on the SDD.


18/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 18/66Syntax-directed Translation Schemes(SDTS) A translation scheme is a context-free grammar in which program fragments called

translation actions are embedded within the right sides of the production.

productions(infix) SDD for postfix notation from infix SDTS for postfix notation from infixlist list + term list.t = list.t || term.t || "+" list list + term {print("+")}

{print("+");} : translation(semantic) action.

1. Design translation schemes(SDTS) for translation

2. Translate :a) parse the input string x andb) emit(execute) the action result encountered during the depth-first traversal of parse

tree.


19/66


Example 2.8.

SDD vs. SDTS for infix to postfix translation.

productions SDD SDTSexpr list + termexpr list + termexpr termterm 0term 1

term 9

expr.t = list.t || term.t || "+"expr.t = list.t || term.t || "-"expr.t = term.tterm.t = "0"term.t = "1"

term.t = "9"

expr list + term printf{"+")}expr list + term printf{"-")}expr termterm 0 printf{"0")}term 1 printf{"1")}

term 9 printf{"0")}

An Example of an Action translation for input 9-5+2

1) Parse.2) Translate.Do we have to maintain the whole parsetree ?No, Semantic actions are performed

during parsing, and we don't need thenodes (whose semantic actions done) anymore.


20/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 20/662.4 PARSINGif token string x L(G), then parse tree

else error message

Top-Down parsing1. At node n labeled with nonterminal A, select one of the productions

whose left part is A and construct children of node n with the symbols on the right side

of that production.

2. Find the next node at which a sub-tree is to be constructed.

ex. G: type simple

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


21/66


An example of top-down parsing for input x=array [ num dotdot num ] of integer


22/66


type simple

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

The selection of production for a nonterminal may involve trial-and-error.

=> backtracking


23/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 23/66Procedure of top-down pars inglet a pointed grammar symbol and pointed input symbol be g, a respectively. if( g N )

then select a production P whose left part equals to g next to current production.( select

expand derivation with that production P.

else if( g = a ) then make g and a be a symbol next to current symbol.else if( g a ) back tracking

let the pointed input symbol a be the symbol that moves back to steps same with thenumber of current symbols of underlying production

eliminate the right side symbols of current production and let the pointed symbol g be theleft side symbol of current production.


24/66


An example of top-down parsing for grammarG : { S->aSb | c | ab }, is acb , aabbL(G)?

S/acbaSb/acbaSb/acb aaSbb/acb X(SaSb) move (SaSb) backtracking

aSb/acbacb/acbacb/acbacb/acb(sc) move move

so, acb L(G)It is finished in 7 steps including one backtracking.

S/aabbaSb/aabbaSb/aabbaaSbb/aabbaaSbb/aabbaaaSbbb/aabb X(SaSb) move (SaSb) move (SaSb) backtracking

aaSbb/aabbaacbb/aabb X(Sc) backtrackingaaSbb/aabbaaabbb/aabb X

(Sab) backtrackingaaSbb/aabb X

backtrackingaSb/aabbacb/aabb

(Sc) bactracking aSb/aabbaabb/aabbaabb/aabbaabb/aabbaaba/aabb

(Sab) move move moveso, aabbL(G)but process is too difficult. It needs 18 steps including 5 backtrackings.


25/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 25/66Predictive parsing (Recursive Decent Parsing, RDP) A strategy for the general top-down parsing

Guess a production, see if it matches, if not, backtrack and try another.

It may fail to recognize correct string in some grammar G and is tedious in processing.

Predictive parsing

is a kind of top-down parsing that

predicts a production whose derived terminal symbol is equal to next input symbol whileexpanding in top-down paring. without backtracking. Procedure decent parser is a kind of predictive parser that is implemented by disjoint

recursive procedures one procedure for each nonterminal, the procedures are patternedafter the productions.


26/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 26/66Procedure of predict ive parsing(RDP)let a pointed grammar symbol and pointed input symbol be g, a respectively. if( g N )

select next production P whose left symbol equals to g and a set of first terminalsymbols of derivation from the right symbols of the production P includes a inputsymbol a.(select

expand derivation with that production P. else if( g = a ) then make g and a be a symbol next to current symbol. else if( g a ) error

=> RDP requires Left Factoring and Elimination of Left-recursion for grammar

G: A a, or

A a 1 1 | a22 | | an n |

for an d


27/66


G : { SaSb | c | ab }According to predictive parsing procedure, acb , aabbL(G)?

S/acb confused in { SaSb, Sab } so, a predictive parser requires some restriction in grammar, that is, there should be only

one production whose left part of productions are A and each first terminal symbol of

those productions have unique terminal symbol.

so G => G1 : { S->aS' | c S'->Sb | ab }

Requirements for a grammar to be suitable for RDP: For each nonterminal either

1. A a, or2. A a11 | a22 | | ann |

for and A may also occur

if none of ai can follow A in a derivation and if we have A

If the grammar is suitable, we can parse efficiently without backtrack.

General top-down parser with backtracking

Recursive(Recursive) Descent Parser without backtracking


28/66


An Example of a Recursive Predictive Parser(RDP) for grammarG : type simple | id | array [ simple ] of typesimple

integer | char | mun dotdot mun procedure match(t: token);

beginif lookahead = t then lookahead = nexttoken

else errorend

procedure type; //cf type simple | id | array [ simple ] of typebegin

if lookahead is in {integer, real, num} then simple //FIRST(simple)={integer,char,num}else if( lookahead = '' then begin match(''); match(id) end //FIRST(id) ={}else if( lookahead = array then begin //FIRST(array [ simple ] of type)={array}

match(array);match('[');simple; match(']');match(of); type endelse error

end procedure simple; //cf simple -> integer | char | mun dotdot mun

beginif lookahead = integer then match(integer)else if lookahead = char then match(char)else if lookahead = num then begin match(num); match(dotdot); match(num) endelse error end;


29/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 29/66Left FactoringIf a grammar contains two productions of form

S a and S a

it is not suitable for top down parsing without backtracking. Troubles of this form can

sometimes be removed from the grammar by a technique called the left factoring.

In the left factoring, we replace { S a, S a } by{ S aS', S' , S' } (Hopefully and start with different symbols)cf. S= a|a=a(|) = aS', S'=|

left factoring for G { SaSb | c | ab }SaS' | c cf. S=aSb | ab | c = a ( Sb | b) | c ) = a S' | c, S'=Sb|bS'Sb | b

A concrete example:

IF THEN |

IF THEN ELSE

is transformed into IF THEN S'

S' ELSE |


30/66


Example,

for G : { SaSb | c | ab }

According to predictive parsing procedure, acb , aabbL(G)?

S/aabb unable to choose { SaSb, Sab ?} According for the feft factored gtrammar G1, acb , aabbL(G)?

G1 : { SaS'|c S'Sb|b}


31/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 31/66Left RecursionAn Example of RDP for a left-recursive Grammar

G: SSa|b, input x=baaS/baaSa/baa Saa/baa Saaa/baa .... (SSa) (SSa) (SSa)

An Example of RDP for a right-recursive GrammarG1 : SbS ' S'aS'|, L(G)=L(G1)S/baabS'/baabS'/baabaS'/baabaS'/baabaaS'/baabaaS'/baabaa/baabaa/baa

(SbS') (move) (S'aS') (move) (S'aS') (move) (S')

A grammar is left recursive iff it contains a nonterminal A, such thatA

+A, where is any string.

Grammar {S S | c} is left recursive because of SS

Grammar {S A, A Sb | c} is also left recursive because of SA Sb

If a grammar is left recursive, you cannot build a predictive top down parser for it.


32/66


33/66


In general, the rule is that

If A A 1 | A2 | | A n and

A 1 | 2 | | m (no i's start with A),then, replace by

A 1R | 2R| | mR and

R 1R | 2R | | nR |

Exercise: Remove the left recursion in the following grammar:

expr expr + term | expr - termexpr term

solution: expr term rest rest + term rest | - term rest |


34/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 34/662.5 A TRANSLATOR FOR SIMPLE EXPRESSIONSConvert infix into postfix(polish notation) using SDT.Abstract syntax (annotated parse tree) tree vs. Concrete syntax tree

Concrete syntax tree : parse tree Fig 2.2. p28.

Abstract syntax tree, syntax tree : Fig 2.20. Concrete syntax : underlying grammar


35/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 35/66Adapting the Translation Scheme Embed the semantic action in the production

Design a translation scheme Left recursion elimination and Left factoring

Example, translate infix expression to post expression

Infix expression system(language)

L={0+1, 2-(5+1), ..}

G=({expr,term},{+,-,0,19},P,expr)

P ={expr expr+ term

expr expr- term

expr term

term 0

term 1

term 9

}

postfix expression system(language)

L'={0 1 +, 2 5 1 + -, ...}

G'=({expr,term},{+,-,0,19},P',expr)

P' = {expr expr term +

expr expr term -

expr term

term 0

term 1

term 9

}


36/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 36/66Example of translator design and executionA translation scheme and with left-recursion.

Fig 2.19. Initial specification for infix-to-postfix translator with left recursion eliminatedexpr expr + term {printf{"+")}expr expr - term {printf{"-")}expr termterm 0 {printf{"0")}term 1 {printf{"1")}

term 9 {printf{"0")}

expr term restrest + term {printf{"+")} restrest - term {printf{"-")} restrest term 0 {printf{"0")}term 1 {printf{"1")}



37/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 37/66Procedure(RDP) for the Nonterminal expr, term, and restm a t c h ( t : t k o e n )

{

i f ( l o o k h e a d = t )

l o o k a h e a d l e x a n ( ) ;

e l s e e r r o r ( ) ;

}

expr term restrest + term {printf{"+")} rest

rest - term {printf{"-")} restrest term 0 {printf{"0")}term 1 {printf{"1")}



38/66


Optimizer and Translator1. expr() {

2. term(); rest();3. }

4. rest() rest()

5. { {

6. if(lookahead == '+' ) { L: if(lookahead == '+' ) {

7. m('+'); term(); p('+'); rest(); m('+'); term(); p('+'); goto L;

8. } else if(lookahead == '-' ) { } else if(lookahead == '-' ) {

9. m('-'); term(); p('-'); rest(); m('-'); term(); p('-'); goto L;

10. } else ; } else ;11. } }

12. expr() {

13. term();

14. while(1) {

15. if(lookahead == '+' ) {

16. m('+'); term(); p('+');

17. } else if(lookahead == '-' ) {

18. m('-'); term(); p('-');19. } else break;

20. }


39/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 39/662.6 LEXICAL ANALYSISreads and converts the input into a stream of tokens to be analyzed by parser.

lexeme : a sequence of characters which comprises a single token.

Lexical Analyzer Lexeme / Token Parser

Removal of White Space and CommentsRemove white space(blank, tab, new line etc.) and commen ts

ContsantsConstants: consider only integers for a while.

eg) for input 31 + 28, output : ?input : 31 + 28output:

tokenattribute, value(or lexeme)


40/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 40/66RecognizingIdentifiers Identifiers are names of variables, arrays, functions... A grammar treats an identifier as a token.

eg) input : count = count + increment;output : ;Symbol table

tokens attributes(lexeme)0

12

3

idid

countincrement

Keywords are reserved, i.e., they cannot be used as identifiers.Then a character string forms an identifier only if it is no a keyword.

punctuation symbols

operators : + - * / := < >


41/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 41/66Interface to lexical analyzer


42/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 42/66A Lexical Analyzer

c=getchcar(); ungetc(c,stdin);

token representation

#define NUM 256

Function lexan()eg) input string 76 + ainput , output(returned value)76 token=NUM tokenval=76 (integer)

+ token=+a toekn=id, tokeval="a"


43/66


A way that parser handles the token NUM returned by laxan()

consider a translation scheme

factor ( expr )| num { print(num.value) }

in the RDP for the above production.

#define NUM 256

...

factor() {

if(lookahead == '(' ) {match('('); exor(); match(')');

} else if (lookahead == NUM) {

printf(" %f ",tokenval); match(NUM);

} else error();

}


44/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 44/66The implementation of function lexan for simple alphabet={integer numbers, +,-,*,/...}

1 ) # include 2) #include

3) in t lino = 1;4) int tokenval = NONE;5) int lexan() {6) int t;7) while(1) {8) t = getchar();9) if ( t==' ' || t=='\t' ) ;10) else if ( t=='\n' ) lineno +=1;11) else if (isdigit(t)) {12) tokenval = t -'0';

13) t = getchar();14) while ( isdigit(t)) {15 ) tokenval = tokenval*10 + t - '0';16) t =getchar();17) }18) ungetc(t,stdin);19) return NUM;20) } else {21 ) tokenval = NONE;22) return t; //+,-,...23) }24) }25) }


45/66


2.7 INCORPORATION A SYMBOL TABLE

The symbol table interfaces, operation, usually called by parser. insert(s,t): input s: lexeme

t: tokenoutput index of new entry

lookup(s): input s: lexemeoutput index of the entry for string s,

or 0 if s is not found in the symbol table.Handling reserved keywords

1. Inserts all keywords in the symbol table in advance.ex) insert("div", div)

token lexeme

insert("mod", mod)

2. while parsing

whenever an identifier s is encountered.

if (lookup(s)'s token is in {keywords} ) s is for a keyword;else s is for a identifier;


46/66


example

preset

insert("div",d iv );insert("mod",m od ); while parsing

count = i + i + div

lookup("count")=>0 insert("count",id );lookup("i") =>0 insert("i",id );lookup("i") =>4, idllokup("div")=>1,d iv


47/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 47/662.8 ABSTRACT STACK MACHINE An abstract machine is for intermediate code generation/execution.

Instruction classes: arithmetic / stack manipulation / control flow3 components of abstract stack machine1) Instruction memory : abstract machine code, intermediate code(instruction)2) Stack 3)Data memory

An example of stack machine operation.

for a input a=11; b=7; (5+a)*b; 1) lexical analysis and parsed(5 a + b *) =>2) intermediate codes : push 5, rvalue 2, +, rvalue 3, *

instruction memory1 push 5 pc2 rvalue 23 +

4 rvalue 3

5 *

Data memory

1 0

2 11 a

3 7 b

Stack

1 5

2 115

3 16

4 716

5 112


48/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 48/66L-value and r-value l-values a : address of location a

r-values a : if a is location, then content of location aif a is constant, then value a

eg) a :=(5 + a ) * b;

lvalue a lvalue 2

r value 5 push 5 r value a rvalue 2 r value of b rvalue 3

Stack ManipulationSome instructions for assignment operation

push v : push v onto the stack.

rvalue a : push the contents of data location a.

lvalue a : push the address of data location a.

pop : throw away the top element of the stack.

:= : assignmen t for the top 2 elements of the stack. copy : push a copy of the top element of the stack.


49/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 49/66Translation of ExpressionsInfix expression(IE) SDD/SDTS Abstact macine codes(AMC) of postfix expression

for stack machine evaluation.eg )

IE: a + b, (PE: a b + ) IC: rvalue arvalue b+

day := (1461 * y) div 4 + (153 * m + 2) div 5 + d( day 1462 y * 4 div 153 m * 2 + 5 div + d + :=)

1) lvalue day 6) div 11) push 5 16) :=2) push 1461 7) push 153 12) div3) rvalue y 8) rvalue m 13) +4) * 9) push 2 14) rvalue d5) push 4 10) + 15) +

A translation scheme for assignment-statement into abstract stack machine code e canbe expressed formally In the form as follows:stmt id := expr

{ stmt.t := 'lvalue' || id .lexeme || expr.t || ':=' }eg) day :=a+b lvalue day rvalue a rvalue b + :=


50/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 50/66Control Flow3 types of jump instructions :

1. Absolute target location

2. Relative target location( distance :Current Target)

3. Symbolic target location( i.e. the machine supports labels)

Control-flow instructions:

l a b e l a : t h e j u m p ' s t a r g e t a

g o t o a : t h e n e x t i n s t r u c t i o n i s t a k e n f r o m s t a t e m e n t l a b e l e d a g o f a l s e a : p o p t h e t o p & i f i t i s 0 t h e n j u m p t o a

g o t r u e a : p o p t h e t o p & i f i t i s n o n z e r o t h e n j u m p t o a h a l t : s t o p e x e c u t i o n


51/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 51/66Translation of Statements Translation scheme for translation if-statement into abstract machine code.

stmt if expr then stmt1{ out := newlabel1)

stmt.t := expr.t || 'gofalse' out || stmt1.t || 'label' out }

Translation scheme for while-statement ?

1) a procedure generates a unique label(eg. zzzzz001, zzzzz002, etc) whenever it is called!


52/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 52/66Emitting a TranslationSemantic Action(Tranaslation Scheme):

stmt if expr { out := newlabel; emit('gofalse', out) } then stmt1 { emit('label', out) }

stmt id { emit('lvalue', id.lexeme) } := expr { emit(':=') }

stmt if expr { out := newlabel; emit('gofalse', out) } then stmt1 { emit('label', out) ; out1 := newlabel; emit('goto', out1); }

else stmt2 { emit('label', out1) ; } if(expr==false) goto out

stmt1

goto out1out : stmt2out1:


53/66


implementation

procedure stmt()

var test,out:integer; begin // stmt id { emit('lvalue', id.lexeme) } := expr { emit(':=') } if lookahead = id then begin emit('lvalue',tokenval); match(id ); match(':='); expr(); emit(':='); end // stmt if expr { out := newlabel; emit('gofalse', out) } then stmt1 { emit('label', out) } else if lookahead = 'if' then begin match('if'); expr(); out := newlabel(); emit('gofalse', out); match('then'); stmt; emit('label', out) end

else error(); end


54/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 54/66Control Flow with Analysisif E1 or E2 then S vs if E1 and E2 then S

E 1 o r E 2 = i f E 1 t h e n t r u e e l s e E 2E 1 a n d E 2 = i f E 1 t h e n E 2 e l s e f a l s e

T h e c o d e fo r E 1 o r E 2 .

Codes for E1 Evaluation result: e1

copy

gotrue OUT

pop

Codes for E2 Evaluation result: e2

label OUT


55/66


T h e f u l l c o d e f o r if E 1 o r E 2 t h e n S ;

codes for E1

copy gotrue OUT1

pop

codes for E2

label OUT1

gofalse OUT2

code for S label OUT2

Exercise: How about if E1 and E2 then S;

if E1 and E2 then S1 else S2;


56/66


2 . 9 P u t t i n g t h e t e c h n i q u e s t o g e t h e r !

infix expression postfix expression

eg) id+(id-id)*num/id id id id - num * id / +

Description of the TranslatorSyntax directed translation scheme(SDTS) to translate the infix expressions

into the postfix expressions, Fig2.35


57/66


Structure of the translator, Fig 2.36

o global header file "header.h"


58/66


59/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 59/66The Parser Module parser.cSDTS Fig 2.35

left recursion elimination

New SDTS Fig 2.38


60/66


An example of parsing for the input 12 div 5 + 2 ;


61/66

Compiler, Computer Eng. Dept., Hanbat National University, Yoon-Joong Kim 61/66The Emitter Module emitter.cemit (t,tval)

The Symbol-Table Modules symbol.c and init.cSymbol.c

data structure of symbol table Fig 2.29 p62

insert(s,t)

lookup(s)

The Error Module error.cExample of execution

input 12 div 5 + 2

output 12

5

div

2

+


62/66



63/66



64/66


65/66



66/66


Documents

2_SimpleOnePassCompiler