pcd lab manual

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GKM College of engineering and technology Allapakam-mappedu road, gkm nagar, Chennai.

Principles of compiler and design – Lab Manual

http://csetube.co.nr/

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IMPLEMENTATION OF A LEXICAL ANALYZER IN C

To write and execute a C program to implement the lexical analyzer.

ALGORITHM:

Start the program.

Declare the file pointer and necessary variables.

Open the input file in the read mode.

Use the analyze function to analyze the input program and store the

identifiers, keywords and operator on idhd, keyhd, ophd respectively.

Stores the tokens in data structure linked lists.

Increment the line number of each token and its occurrences.

Using the show function print the linked lists in a tabular format.

Stop the program.

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PROGRAM #include<stdio.h>

#include<ctype.h> #include<string.h>

#include<malloc.h> struct lineno {

int ln; struct lineno *next;

}; struct lx {

char str[20]; struct lineno *next;

struct lx *down; }; typedef struct lineno node;

typedef struct lx lex; lex *keyhd,*ophd,*idhd;

FILE *fp; char temp[20]; int line=1;

lex *getkeynode(); node *getlnode();

void analyze(); void add(lex *); void show(lex *);

int main() {

keyhd=getkeynode(); ophd=getkeynode(); idhd=getkeynode();

fp=fopen("inp.c","rt"); if(fp==NULL)

{ printf("No input file!!"); /*exit(1);*/

} analyze();

printf("\n\t\tKeywords\n\n"); show(keyhd); printf("\n\t\tIDENTIFIER\n\n");

show(idhd); printf("\n\t\tOPERATORS\n\n");

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show(ophd); free(ophd);

free(idhd); free(keyhd);

return 0; } lex *getkeynode()

{ lex *temp;

temp=(lex*)malloc(sizeof(struct lx)); temp->next=NULL; temp->down=NULL;

return temp; }

node *getlnode() { node *temp;

temp=(node*)malloc(sizeof(struct lineno)); temp->next=NULL;

return temp; } void analyze()

{ char ch;

int i,flag; ch=fgetc(fp); while(ch!=EOF)

{ if(ch=='\n')

{ ch=fgetc(fp); line++;

continue; }

if(isspace(ch)) { ch=fgetc(fp);

continue; }

if(ch== '/') while((ch=fgetc(fp))!='/'); if(isalpha(ch))

{ i=0;

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temp[i++]=ch; ch=fgetc(fp);

} while(isalpha(ch)||ch=='.');

fseek(fp,-1,1); temp[i]='\0'; i=0;

flag=0; while(strlen(key[i]))

{ if(strcmp(temp,key[i])==0) {

add(keyhd); flag=1;

break; } i++;

} if(flag==0)

add(idhd); } else

{ temp[0]=ch;

temp[1]='\0'; add(ophd); }

ch=fgetc(fp); }

} void add(lex *hd) {

node *l,*m; lex *q,*p=hd->down;

while(p) { if(strcmp(p->str,temp)==0)

break; p=p->down;

} if(p==NULL) {

p=hd; while(p->down)p=p->down;

q=getkeynode(); p->down=q;

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strcpy(q->str,temp); l=getlnode();

l->ln=line; q->next=l;

} else {

m=getlnode(); l=p->next;

m->ln=line; if(l==NULL) p->next=m;

else {

while(l->next) l=l->next; l->next=m;

} void show(lex *hd) {

lex *p=hd->down,*q; node *l,*m;

int i=1,occ=0; printf("\tTokens\t\tLineno\t\tOccurences\n\n"); while(p)

{ occ=1;

l=p->next; printf("\t%s\t\t",p->str); printf("%d",l->ln);

l=l->next; while(l)

{ printf("%d",l->ln); l=l->next;

occ++; }

printf("\t\t%d",occ); p=p->down; printf("\n");

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INPUTFILE: #include<stdio.h>

void main() {

int a,b,c; printf("Enter the value of a& b");

scanf("%d%d",&a,&b); c=a+b;

printf("%d",c); }

OUTPUT: [3cse09@localhost ~]$ cc exp1.c

[3cse09@localhost ~]$ ./a.out KEYWORDS

Tokens Lineno Occurences

include 1 1 stdio.h 1 1

void 2 1 main 2 1

int 4 1 printf 5,8 2 scanf 6 1

IDENTIFIER

Tokens Lineno Occurences

a 4,5,6,7 4 b 4,5,6,7 4

c 4,7,8 3 Enter 5 1 the 5 1

value 5 1 of 5 1

d 6,6,8 3

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OPERATORS

Tokens Lineno Occurences # 1 1

< 1 1 > 1 1

( 2,5,6,8 4 ) 2,5,6,8 4 { 3 1

, 4,4,6,6,8 5 ; 4,5,6,7,8 5

" 5,5,6,6,8,8 6 & 5,6,6 3 % 6,6,8 3

= 7 1 + 7 1

RESULT: Thus the implementation of lexical analyzer was successfully done.

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IMPLEMENTATION OF A LEXICAL ANALYZER USING LEX TOOL

To implement the lexical analyzer using lex tool for a subset of C language.

ALGORITHM:

Start the program.

Declare necessary variables and creates token representation using

Regular.

Print the pre processor or directives, keywords by analysis of the input

program.

In the program check whether there are arguments.

Declare a file and open it as read mode.

Read the file and if any taken in source program matches with RE that all

returned as integer value.

Print the token identified using YYdex() function.

Stop the program.

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PROGRAM:

%{ %} identifier[a-zA-Z][a-zA-Z0-9]*

%% #.* {printf("\n%s is a preprocessor directive",yytext);}

int | float | char |

double | while |

do | if | break |

continue | void |

switch | return | else |

goto {printf("\n%s is a keyword",yytext);} {identifier}\( {printf("\n function %s",yytext);}

\{ {printf("\nblock begins");} \} {printf("\nblock ends");} \( {printf("\n");ECHO;}

{identifier}(\[[0-9]*\])* {printf("\n%s is an identifier",yytext);} \".*\" {printf("\n %s is a string ",yytext);}

[0-9]+ {printf("\n%s is a number",yytext); } \<= |

\>= | \< |

\> | \== {printf("\n %s is a relational operator",yytext);} \= |

\+ | \- |

\/ | \& | % {printf("\n %s is a operator",yytext);}

. | \n;

%% int main(int argc,char **argv)

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{ FILE *file;

file=fopen("inp.c","r"); if(!file)

{ printf("could not open the file!!!"); exit(0);

} yyin=file;

yylex(); printf("\n\n"); return(0);

} int yywrap()

{ return 1; }

INPUT FILE:

#include<stdio.h> void main()

{ int a,b,c; printf("enter the value for a,b");

scanf("%d%d",&a,&b)'; c=a+b;

printf("the value of c:%d",&c); }

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OUTPUT: [3cse01@localhost ~]$ lex ex3.l

[3cse01@localhost ~]$ cc lex.yy.c [3cse01@localhost ~]$ ./a.out

#include<stdio.h> is a preprocessor directive void is a keyword

function main( block begins

int is a keyword a is an identifier b is an identifier

c is an identifier function printf(

"enter the value for a,b" is a string function scanf( "%d%d" is a string

& is a operator a is an identifier

& is a operator b is an identifier c is an identifier

= is a operator a is an identifier

+ is a operator b is an identifier function printf(

"the value of c:%d" is a string & is a operator

c is an identifier block ends

RESULT: Thus the program to implement the lexical analyzer using lex tool for a subset of

C language was implemented and verified.

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IMPLEMENTATION OF A RECURSIVE DESCENT PARSER

To implement a recursive descent parser in a C program.

ALGORITHM:

Start the program.

Get the expression from the user and call the parser() function.

In lexer() get the input symbol and match with the look ahead pointer and then return the token accordingly.

In E(), check whether the look ahead pointer is ‘+’ or ‘-‘ else

return syntax error.

In T(),check whether the look ahead pointer is ‘*’ or ‘/’ else

return syntax error.

In F(),check whether the look ahead pointer is a member of any

identifier.

In main(), check if the current look ahead points to the token in a given CFG it doesn’t match the return syntax error.

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PROGRAM: #include<stdio.h> #include<string.h>

#include<stdlib.h> #define DONE 260 #define NUM 257

#define ID 259 void E();

void T(); void F(); void parser();

int lexer(); void match(int);

void parsex(); int i=0,flag=9,lkahd; char inp[75];

int cnt=-1; int main()

{ int i=-1; char c;

FILE *fp; fp=fopen("ep1.text","rt");

while((c=fgetc(fp))!=EOF) { inp[i++]=c;

} inp[i]=EOF;

parser(); } int lexer()

{ int t;

while(1) { t=inp[cnt++];

if(t==' '||t=='\t'||t=='\n') ;

else if(t=='+'||t=='-'||t=='*'||t=='/') { printf("arithmetic operator %c\n",t);

return t; }

else if(isdigit(t)) {

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printf("\n number: %c\n",t); return NUM;

} else if(isalpha(t))

{ printf("\n identifier: %c\n",t); return ID;

} else if(t==EOF)

return DONE; else return t;

}} void parser()

{ lkahd=lexer(); while(lkahd!=DONE)

{ E();

match(';'); } if(lkahd==DONE)

parsex(); }

void match(int t) { if(lkahd==t)

lkahd=lexer(); else

return; } void F()

{ switch(lkahd)

{ case '(': match('(');

E(); match(')');

break; case NUM: match(NUM);

break; case ID:

match(ID); break;

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default:{ printf("syntax error");

flag=7; }}}

void T() { int t;

F(); while(1)

{ switch(lkahd) {

case '*': t=lkahd;

match(lkahd); F(); continue;

case '/': t=lkahd;

match(lkahd); F(); continue;

default: return;

}}} void E() {

int t; T();

while(1) { switch(lkahd)

{ case '+':

t=lkahd; match(lkahd); T();

continue; case '-':

t=lkahd; match(lkahd); T();

continue; default:

return; }}}

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void parsex() {

if(flag!=7) printf("parse seccessfull\n");

else printf("parse successfull\n errors found\n"); exit(0);

INPUT FILE: (a+b)*(c/d)-g;

1+5; 5-3; a+2;

OUTPUT: [3cse01@localhost ~]$ cc ex4.c [3cse01@localhost ~]$ ./a.out identifier: a

arithmetic operator + identifier: b

arithmetic operator * identifier: c arithmetic operator /

identifier: d arithmetic operator -

identifier: g number: 1 arithmetic operator +

number: 5 number: 5

arithmetic operator - number: 3 identifier: a

arithmetic operator + number: 2

parse seccessfull

RESULT: Thus the program to implement the recursive descent parser was implemented and

verified.

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CONVERSION OF RE TO NFA

To write and execute a C program to convert RE to NFA.

ALGORITHM:

Start the program.

Get the RE and store it in RE.

Using a loop check whether the RE contains (re, [1], a) if so only a

function.

If re[i] contain b, then call on bfn.

Then check whether re[i] is ‘*’ (or) ‘/’.

If there is ‘*’ call star ().

If there is ‘/’then call or().

Continue the NFA’s of all the component expression into single NFA and

print the corresponding RE.

Stop the program.

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PROGRAM: #include<stdio.h> #include<string.h>

char re[35]; int sta[20],fna[20],stb[20],fnb[20],ste[20],fne[20],cnt=0,cnta=0,cntb=0;

int cnte=0,tag=0,i=0,j=0,k=0,f=0,a=0,b=0,a1=0,b1=0; void onlya(int sa,int fa) {

sta[cnta]=sa; fna[cnta]=fa;

cnta++; } void onlyb(int sa,int fa)

{ stb[cntb]=sa;

fnb[cntb]=fa; cntb++; }

void onlye(int sa,int fa) {

ste[cnte]=sa; fne[cnte]=fa; cnte++;

} void star() {

int p=0; if(re[i-1]!=')')

{ if(re[i-1]=='a') {

onlye(fna[cnta-1],sta[cnta-1]); onlye(sta[cnta-1],fna[cnta-1]);

} if(re[i-1]=='b') {

onlye(stb[cntb-1],fnb[cntb-1]); onlye(fnb[cntb-1],stb[cntb-1]);

}} else {

j=i; do

{ j--;

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if(re[j]=='a') a1++;

if(re[j]=='b') b1++;

if(re[j]==')') p++; if(re[j]=='(')

p--; }

while((re[j]!='(')||(p>0)); if((re[j+1]=='a')||(re[j+2]=='a')) {

onlye(cnt,sta[cnta-a1]); onlye(sta[cnta-a1],cnt);

} if((re[j+1]=='b')||(re[j+2]=='b')) {

onlye(cnt,stb[cntb-b1]); onlye(stb[cntb-b1],cnt);

}}} void or() {

if((re[i-1]!='a')||(re[i-1]!='b')) {

for(k=i-1;k>=0;k--) { if(re[k]=='a')

a++; if(re[k]=='b')

b++; if(re[k]=='(') {

onlye(++cnt,cnta-a); if(re[k+1]=='b')

onlye(++cnt,cntb-b); } }

if(re[i-1]=='(') i++;

if(re[i+1]=='a') { i++;

f=cnt; onlye(f,++cnt);

onlya(f,++cnt); }

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if(re[i+1]=='b') {

i++; f=cnt;

onlye(f,++cnt); onlyb(f,++cnt); }

} else

{ if((re[i-1]=='a')&&(re[i+1]=='b')) {

onlyb(sta[cnta-1],++cnt); onlye(fna[cnta-1],++cnt);

onlye(fnb[cntb-1],cnt); i++; }

if((re[i-1]=='b')&&(re[i+1]=='a')) {

onlya(stb[cntb-1],++cnt); onlye(fnb[cntb-1],++cnt); onlye(fna[cnta-1],cnt);

i++; }}}

int main() { printf("Enter the expression:");

scanf("%s",re); for(i=0;i<strlen(re);i++)

{ if(re[i]=='a') {

f=cnt; onlya(f,++cnt);

} if(re[i]=='b') {

f=cnt; onlyb(f,++cnt);

} if(re[i]=='*') star();

if(re[i]=='|') or();

} printf("\n states\t|a\t|b\t|e\n");

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printf("......................................."); while(tag<=cnt)

{ printf("\n");

printf("{%d}\t|",tag); for(k=0;k<cnta;k++) if(tag==sta[k])

printf("{%d}",fna[k]); putchar('\t');

putchar('|'); for(k=0;k<=cntb;k++) if(tag==stb[k])

printf("{%d}",fnb[k]); putchar('\t');

putchar('|'); for(k=0;k<cnte;k++) if(tag==ste[k])

printf("{%d}",fne[k]); tag++;

printf("\n\t|\t|\t|"); } putchar('\n');

return 0; }

OUTPUT: [2cse54@localhost ~]$ ./a.out

Enter the expression:(ab)*

states |a |b |e ....................................... {0} |{1} |{0} |{2}

| | | {1} | |{2} |{2}

| | | {2} | | |{0}{1} | | |

RESULT: Thus the program to convert RE to NFA was implemented and verified.

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ANALYSIS OF SOURCE PROGRAM USING LEX AND YAC TOOL

To write lex and yac program for implementation of syntax analyzer using lex and yac tool.

ALGORITHM:

Start the program.

In lex program start lex specification with declararive section.

In translating rule section define data type identifier along with

their action.

In main function open the input file in read mode.

The key words, identifiers and data type are defined.

Stop the program.

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PROGRAM:

%{ /*Definition Section*/

#include<stdio.h> #include"y.tab.h"

//to keep track of errors int lineno=1; %}

identifier[a-zA-Z][a-zA-Z]* number[0-9]+

%% main|(1)return MAIN; if return IF;

else return ELSE; while return WHILE;

int | char | float |

long return TYPE; {identifier} return VAR;

{number} return NUM; \< | \> |

\<= | \>= |

==return RELOP; [:space:]+; [\t]+;

\n lineno++; .return yytext[0];

%% int yywrap() {

return 1; }

Exp5.y:

%{ #include<stdio.h>

#include<string.h> extern int lineno;

int errno=0; %}

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%token NUM VAR RELOP MAIN %token IF ELSE WHILE TYPE

//define precedence and associativity of operators] %left'-''+'

%left'*''/' %% PROGRAM:MAIN BLOCK

; BLOCK:'{'CODE'}'

; CODE:BLOCK |STATEMENT CODE

|STATEMENT ;

STATEMENT:DECST';' |DECST{printf("\n missing ';' lineno%d",lineno);errno++;} |ASSIGNMENT';'

|ASSIGNMENT {printf("\nMissing ';' lineno%d",lineno);errno++;} |CONDST

|WHILEST ; DECST:TYPE VARLIST

; VARLIST:VAR','VARLIST

|VAR ; ASSIGNMENT:VAR'='EXPR

; EXPR:EXPR'+'EXPR

|EXPR'-'EXPR |EXPR'*'EXPR |EXPR'/'EXPR

|'-'EXPR |'('EXPR')'

|VAR |NUM ;

CONDST:IFST |IFST ELSEST

; IFST:IF'('CONDITION')' BLOCK

; ELSEST:ELSE

BLOCK ;

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CONDITION:VAR RELOP VAR |VAR RELOP NUM

|VAR |NUM

; WHILEST:WHILELOOP ;

WHILELOOP:WHILE'('CONDITION')' BLOCK

; %% #include"lex.yy.c"

extern FILE *yyin; int main(int argc,char *argv[])

{ FILE *fp; int i;

fp=fopen(argv[1],"r"); yyin=fp;

while(!feof(yyin)) {

yyparse(); }

if(errno==0) printf("\n no error found!! \n parsing successfull\n"); else

printf("\n%derrno(s) found!!",errno); putchar('\n');

return 0; } yyerror()

{ printf("\n error on lineno:%d\n",lineno);

errno++; }

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Input:

main() {

a=5; b=4;

c=a+b; if(a>b) {

a=1; }

else { c=2;

} while(c==1)

{ a=1; }

Output:

[2cse54@localhost]lex exp5.l [2cse54@localhost]yacc -d exp5.y

[2cse54@localhost]cc y.tab.c -ll

No Errors Found!! Parsing Sucessfull

RESULT:

Thus the program for implementation of a syntax analyzer using lex

and yac tools was successfully done.

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ANALYSIS OF SOURCE PROGRAM USING LEX AND YAC TOOL

To write lex and yac program for implementating calculator using lex

and yac tool. ALGORITHM:

Start the program.

In lex program declare the identifier for log, cos, sin, tan and

memory.

Identify the identifier and return id to parser.

In yac program declare the possible symbol type which are the

tokens which are returned by lex.

Define precedence and associativity.

Define rule in cfg for non terminal.

In main() get the expression from user and print the output.

Stop the program.

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PROGRAM: %{ #include"y.tab.h"

#include<math.h> %}

%% ([0-9]+|([0-9]*\.[0-9]+)([eE][-+]?[0-9]+)?) {yylval.dval=atof(yytext);return NUMBER;}

log | LOG {return LOG;}

In {return nLOG;} sin | SIN {return SINE;}

cos | COS {return COS;}

tan | TAN {return TAN;} mem {return MEM;}

[\t]; \$ return 0;

\n|. return yytext[0]; %% Exp6.y:

%{ double memvar; %}

%union {

double dval; } %token<dval>NUMBER

%token<dval>MEM %token LOG SINE nLOG COS TAN

%left'-''+' %left'*''/' %right'^'

%left LOG SINE nLOG COS TAN %nonassoc UMINUS

%type<dval>expression %% start:statement'\n'

|start statement'\n' ;

statement:MEM'='expression {memvar=$3;} | expression{printf("Answer=%g\n",$1);}

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; expression:expression'+'expression {$$=$1+$3;}

| expression '-' expression {$$=$1-$3;} | expression '*' expression {$$=$1*$3;}

| expression '/' expression { if($3==0)

yyerror("divide by zero"); else

$$=$1/$3; } |expression'^'expression {$$=pow($1,$3);}

; expression:'-'expression %prec UMINUS{$$=-$2;}

|'('expression')'{$$=$2;} |LOG expression {$$=log($2)/log(10);} |nLOG expression {$$=log($2);}

|SINE expression {$$=sin($2*3.14/180);} |COS expression {$$=cos($2*3.14/180);}

|TAN expression {$$=tan($2*3.14/180);} |NUMBER {$$=$1;} |MEM {$$=memvar;}

main() { printf("Enter the expression");

yyparse();} int yyerror(char *error)

{ printf("%s\n",error);}

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Output:

[2cse54@localhost]$lex exp6.l [2cse54@localhost]$yacc -d exp6.y

[2cse54@localhost]$cc lex.yy.c y.tab.c -ll -lm [2cse54@localhost]$./a.out Enter the expression:(5+2)*(3-1)/(2)

Answer=7 mem=cos45

sib45/mem Answer=1

RESULT:

Thus the program for implementating calculating using lex and yac

tools was successfully done.

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CONSTRUCTION OF TOP DOWN PARSING TABLE

To write a c program that implement the construction of top-down parsing table.

ALGORITHM:

Start the program.

Initial get the number of terminals and non-terminals separately present in the

string from the user.

After entering terminals and non-terminals call function such that each terminal with some non-terminals as the input gives out either or own terminal or no value.

Generate the table accordingly by the function called inside the program .

With the help of the table check the string whether it is valid or not.

Terminate the program.

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PROGRAM:

#include<stdio.h>

#include<string.h> #include<ctype.h>

#include<stdlib.h> int n,m,j,i,k,q,l,kl,ptr=0,x=0; char nt[10],t[10],top[10][10][10],in[10],stack[10],temp[5],temp2[10];

void terminal(void); void nonterminal(void);

void strrev(char *str); int main() {

int f,f1; stack[ptr]='$';

printf("enter the no.of terminals"); scanf("%d",&m); printf("enter the terminals\n");

printf("--------------------\n"); for(i=0;i<m;i++)

scanf("%s",&t[i]); printf("enter the no.of non terminals"); scanf("%d",&n);

printf("enter the non terminals\n"); printf("---------------------\n");

for(i=0;i<n;i++) scanf("%s",&nt[i]); for(i=0;i<n;i++)

{ for(j=0;j<m;j++)

{ printf("enter the production for[%c][%c]::>",nt[i],t[j]); scanf("%s",top[i][j]);

printf("------------------------\n"); for(j=0;j<m;j++)

printf("\t%c",t[j]); printf("\n"); for(q=0;q<n;q++)

{ printf("\n");

printf("%c",nt[q]); for(j=0;j<m;j++) {

if(strcmp(top[q][j],"0")==0) printf("\t-");

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else printf("\t%c->%s",nt[q],top[q][j]);

printf("\nenter the string:"); scanf("%s",in); l=strlen(in);

i=0; stack[++ptr]=nt[0];

printf("\n stack\t\t input \t\taction\n"); printf("%s\t\t%s\t\t",stack,in); do

{ k=kl=0;

if((in[i]==stack[ptr])&&(in[i]=='$')&&(stack[ptr]=='$')) { printf("\t \tvalid string");

exit(0); }

else if((in[i]==stack[ptr])&&(in[i]!='$')&&(stack[ptr]!='$')) { printf("\t \ttop %c off the stack\n",stack[ptr]);

stack[ptr--]='\0'; for(f=0;f<=ptr;f++)

printf("%c",stack[f]); printf("\t\t"); for(f=i+1;f<l;f++)

printf("%c",in[f]); i++;

} else {

if(isupper(stack[ptr])) {

nonterminal(); terminal(); if(strcmp(top[k][kl],"e")==0)

stack[ptr--]='\0'; else if(strcmp(top[k][kl],"0")==0)

{ printf("\t\tINVALID STRING"); exit(1);

} else

{ strcpy(temp,top[k][kl]);

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int b=0; int v=0;

for(v=strlen(temp)-1;v>=0;v--) temp2[b++]=temp[v];

temp2[b]='\0'; strcpy(temp,temp2); stack[ptr--]='\0';

strcat(stack,temp); ptr=strlen(stack)-1;

} printf("\t\tpushing.. %c->%s\n",nt[k],top[k][kl]); for(f=0;f<=ptr;f++)

printf("%c",stack[f]); printf("\t\t");

for(f=x;f<l;f++) printf("%c",in[f]); x++;

}}} while(ptr>=0);

return 0; } void nonterminal()

{ for(k=0;k<n;k++)

if(nt[k]==stack[ptr]) break; }

void terminal() {

for(kl=0;kl<m;kl++) if(t[kl]==in[i]) break;

OUTPUT:

enter the no.of terminals5 enter the terminals

-------------------- a

$ enter the no.of non terminals3

enter the non terminals ---------------------

http:/

/csetu

A enter the production for[S][a]::>a

enter the production for[S][(]::>(L) enter the production for[S][)]::>0 enter the production for[S][,]::>0

enter the production for[S][$]::>0 enter the production for[L][a]::>SA

enter the production for[L][(]::>SA enter the production for[L][)]::>0 enter the production for[L][,]::>0

enter the production for[L][$]::>0 enter the production for[A][a]::>0

enter the production for[A][(]::>0 enter the production for[A][)]::>e enter the production for[A][,]::>SA

enter the production for[A][$]::>0

parser table------------------------ a ( ) , $

S S->a S->(L) - - - L L->SA L->SA - - -

A - - A->e A->SA - enter the string:(a,a)$

Stack Input Action

$S (a,a)$ pushing.. S->(L) $)L( (a,a)$ top ( off the stack

$)L a,a)$ pushing.. L->SA $)AS a,a)$ pushing.. S->a

$)Aa ,a)$ top a off the stack $)A ,a)$ pushing.. A->SA $)AS a)$ INVALID STRING

RESULT:

Thus the program to construct a top down parsing table was successfully done.

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/csetu

GENERATION OF INTERMEDIATE CODE USING LEX AND YAC TOOLS

To write a program in linux to generate intermediate code using LEX and YACC.

ALGORITHM:

Start the program.

Include the header file.

In int code.l,declare the variable lie no as integer and assign it to be equal to ‘1’.

Start the int code.l with declarative section.

In translation rules section define keywords ,data types and integer along with

their actions .

Start the main block.

In main block check the statement

1.declarative 2.assignment 3.conditional 4.if and else 5.While assignment.

Perform the actions of that particular block.

In main program declare the parameters arg c as int end *argv[] as char.

In main program open file in read mode.

Print the output in a file.

End the program.

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/csetu

PROGRAM:

#include"y.tab.h" #include<stdio.h>

int LineNo=1; %} identifier[a-zA-Z][_a-zA-Z0-9]*

number[0-9]+([0-9]*\.[0-9]+) %%

Exp.y: %{ #include<string.h>

#include<stdio.h> struct quad

{ char op[5]; char arg1[10];

char arg2[10]; char result[10];

}QUAD[30]; struct stack {

int item[100]; int top; }stk;

int Index=0,tIndex=0,StNo,Ind,tInd; extern int LineNo;

%} %union {

char var[10]; }

PROGRAM: MAIN BLOCK ; BLOCK: '{' CODE '}'

CODE: BLOCK | STATEMENT CODE

| STATEMENT ; STATEMENT: DECST';'

| ASSIGNMENT ';' | CONDST

| WHILEST ;

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/csetu

DECST: TYPE VARLIST ;

VARLIST: VAR ',' VARLIST | VAR

; ASSIGNMENT: VAR '='EXPR { strcpy(QUAD[Index].op,"=");

strcpy(QUAD[Index].arg1,$3); strcpy(QUAD[Index].arg2,"");

strcpy(QUAD[Index].result,$1); strcpy($$,QUAD[Index++].result); }

; EXPR: EXPR '+' EXPR {AddQuadruple("+",$1,$3,$$); }

| EXPR '-' EXPR {AddQuadruple('-',$1,$3,$$); } | EXPR '*' EXPR {AddQuadruple('*',$1,$3,$$); } | EXPR '/' EXPR {AddQuadruple('/',$1,$3,$$); }

| '-' EXPR {AddQuadruple("UMIN",$2,"",$$); } | '(' EXPR ')' {strcpy($$,$2); }

| VAR | NUM ;

CONST:IFST{ Ind=pop();

sprintf(QUAD[Ind].result,"%d",Index); Ind=pop(); sprintf(QUAD[Ind].result,"%d",Index);

} | IFST ELSEST

; IFST:IF '('CONDITION')' {

strcpy(QUAD[Index].op,"=="); strcpy(QUAD[Index].arg1,$3);

strcpy(QUAD[Index].arg2,"FALSE"); strcpy(QUAD[Index].result,"-1"); push(Index);

Index++; }

BLOCK{ strcpy(QUAD[Index].op,"GOTO"); strcpy(QUAD[Index].arg1,"");

strcpy(QUAD[Index].arg2,""); strcpy(QUAD[index].result,"-1");

push(Index); Index++;

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/csetu

ELSEST:ELSE{ tInd=pop();

Ind=pop(); push(tInd); sprintf(QUAD[Ind].result,"%d",Index);

} BLOCK

{ Ind=pop(); sprintf(QUAD[Ind].result,"%d",Index);

AddQuadruple($2,$1,$3,$$); StNo=Index-1; |NUM

; WHILEST:WHILELOOP{

Ind=pop(); sprintf(QUAD[Ind].result,"%d",StNo); Ind=pop();

sprintf(QUAD[Ind].result,"%d",Index); }

; WHILELOOP:WHILE '('CONDITION')' { strcpy(QUAD[Index].op,"==");

strcpy(QUAD[Index].arg1,$3); strcpy(QUAD[Index].arg2,"FALSE");

strcpy(QUAD[Index].result,"-1"); push(Index); Index++;

} BLOCK {

strcpy(QUAD[Index].op,"GOTO"); strcpy(QUAD[Index].arg1,""); strcpy(QUAD[Index].arg2,"");

strcpy(QUAD[index].result,"-1"); push(Index);

Index++; } %%

extern FILE *yyin; int main(int argc,char *argv[])

{ FILE *fp;

http:/

/csetu

int i; if(argc>1)

{ fp=fopen(argv[1],"r");

if(!fp) { printf("\n file not found!");

exit(0); }

yyin=fp; } yyparse();

printf("\n\n \t\t............\n\t\t pos operator arg1 arg2 result \n\t\t....."); for(i=0;i<Index;i++)

{ printf("\n\t\t%d\t%s\t%s\t%s\t%s",i,QUAD[i],op,QUAD[i].arg1,QUAD[i].arg2,QUAD[i] .result);

} printf("\n\t--------------");

printf("\n\n"); return 0; }

void push(int data) {

stk.top++; if(stk.top==100) {

printf("\nstack overflow"); exit(0);

} stk.items[stk.top]=data; }

int pop(){ int data;

if(stk.top==-1) { printf("\nstack underflows");

exit(0);} data=stk.items[stk.top--];

return data; } void AddQuadruple(char op[5],char arg1[10],char arg2[10],char result[10])

{ strcpy(QUAD[Index].op,op);

strcpy(QUAD[Index].arg1,arg1); strcpy(QUAD[Index].arg2,arg2);

http:/

/csetu

sprintf(QUAD[Index].result,"t%d",tIndex++); strcpy(result,QUAD[Index++].result);

} yyerror()

{ printf("\nError on line no: %d",LineNo); }

Input:

main() { if(a<b)

{ a=a+b;

} while(a<b) {

a=a+b; }

} Output:

[2cse54@localhost]$./a.out in.c

....................................................................................... Pos Operator A Arg2 Result

0 < a b t0 1 == t0 FALSE 5 2 + a b t1

3 = t1 a 4 GOTO 5

5 < a b t2 6 == t2 FALS t0 7 + a b t3

8 = t3 a ...........................................................................................

RESULT: Thus the program to generate an intermediate code using lex and yac

tools was successfully done

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/csetu

COVERSION OF INTERMEDIATE CODE INTO ASSEMBLY LANGUAGE INSTRUCTION

AIM: To convert the given intermediate Code into assembly language instruction using

c program.

ALGORITHM:

Start the program.

Open the source file and store the contents as quadruples.

Check for operators, in quadruples, if it is an arithmetic operator generate it or if an assignment operator generate it, else perform unary minus on register c.

Write the generated code into output definition of the file.

Print the output.

Stop the program.

http:/

/csetu

PROGRAM :

#include<stdio.h>

#include<string.h> #include<stdlib.h>

struct quad { char op[10],arg1[10],arg2[10],result[10];

}QUAD[30]; FILE *src,*dest;

int Reg_Index=0; void Assignment(); void Un_Minus();

void Gen_Code(char op[5]); int i=0,j=0;

void getquad() { src=fopen("inp.c","r");

while(!feof(src)) {

fscanf(src,"%s",QUAD[i].op); fscanf(src,"%s",QUAD[i].arg1); fscanf(src,"%s",QUAD[i].arg2);

fscanf(src,"%s",QUAD[i].result); i++;

} } int main()

{ getquad();

dest=fopen("outp.c","w"); while(j<i) {

if(!strcmp(QUAD[j].op,"+")) Gen_code("ADD");

if(!strcmp(QUAD[j].op,"=")) Assignment(); if(!strcmp(QUAD[j].op,"-"))

Gen_code("SUB"); if(!strcmp(QUAD[j].op,"*"))

Gen_code("MUL"); if(!strcmp(QUAD[j].op,"/")) Gen_code("DIV");

if(!strcmp(QUAD[j].op,"UMIN")) Un_Minus();

j++; }

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/csetu

fcloseall(); return 0;

Un_Minus(); j++;

} fcloseall(); return 0;

} void Gen_code(char op[5])

{ char strIns[25]; sprintf(strIns,"MOV R%d,%s\n",Reg_Index,QUAD[j].arg1);

fwrite(&strIns,strlen(strIns),1,dest); sprintf(strIns,"%s R%d,%s\n",op,Reg_Index++,QUAD[j].arg2);

fwrite(&strIns,strlen(strIns),1,dest); { fscanf(src,"%s",QUAD[i].op);

fscanf(src,"%s",QUAD[i].arg1); fscanf(src,"%s",QUAD[i].arg2);

fscanf(src,"%s",QUAD[i].result); i++; }

} int main()

{ getquad(); dest=fopen("outp.c","w");

while(j<i) {

if(!strcmp(QUAD[j].op,"+")) Gen_code("ADD"); if(!strcmp(QUAD[j].op,"="))

Assignment(); if(!strcmp(QUAD[j].op,"-"))

Gen_code("SUB"); if(!strcmp(QUAD[j].op,"*")) Gen_code("MUL");

if(!strcmp(QUAD[j].op,"/")) Gen_code("DIV");

if(!strcmp(QUAD[j].op,"UMIN")) Un_Minus(); j++;

} fcloseall();

return 0; }

http:/

/csetu

void Gen_code(char op[5]) {

char strIns[25]; sprintf(strIns,"MOV R%d,%s\n",Reg_Index,QUAD[j].arg1);

fwrite(&strIns,strlen(strIns),1,dest); sprintf(strIns,"%s R%d,%s\n",op,Reg_Index++,QUAD[j].arg2); fwrite(&strIns,strlen(strIns),1,dest);

} void Assignment()

{ char strIns[25]; if(QUAD[j].arg1[0]!='t')

{ sprintf(strIns,"MOV R%d,%s\n",Reg_Index++,QUAD[j].arg1);

fwrite(&strIns,strlen(strIns),1,dest); } sprintf(strIns,"MOV %s,R%d\n",QUAD[j].result,Reg_Index-1);

fwrite(&strIns,strlen(strIns),1,dest); }

void Un_Minus() { char strIns[25];

sprintf(strIns,"MOV R%d,o\n",Reg_Index); fwrite(&strIns,strlen(strIns),1,dest);

sprintf(strIns,"SUB R%d,%s\n",Reg_Index++,QUAD[j].arg1); fwrite(&strIns,strlen(strIns),1,dest); }

Input: inp.c

+ a b t0

- t0 NULL a ' c a t1

= t1 NULL a / c a t2 = t2 NULL b

= a NULL c

http:/

/csetu

OUTPUT: outp.c

MOV R0,a

ADD R0,b MOV R1,t0 SUB R1,NULL

MOV a,R1 MOV R2,c

DIV R2,a MOV b,R2 MOV R3,a

MOV c,R3

RESULT:

Thus the program to convert the given intermediate Code into assembly language instruction using c program was successfully done.

pcd lab manual

recursive descent parser

return syntax error

include void main

identifier function printf

main block

int main

lexical analyzer

3cse09 localhost

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