CC

COMPILER CONSTRUCTIONS LAB RECORD INDEXS.NO LIST OF PROGRAMS PAGE NO.

1 Program to implement a standalone scanner.

2 Write a program in lex to check if a no is octal, Hexadecimal

3 Program in lex to capitalize all comments in a c program

4 Program to implement a lex scanner

5 Write a program to find first and follow elements

6 Write a Program to find canonical LR(0) collections. i.e. a). closure

7 Write a Program to construct SLR parsing table using program 6

8 Write a program to implement a calculator using yacc. a). using lex b). without using lex

9 Program to generate code

10 Program to optimize the code

1. Program to implement a standalone scanner.

Scanner.c#include <stdio.h>#include <string.h>#include <ctype.h>

main(){

FILE *in,*out;char str1[20],str2[20];char ch;int k,flag;char key_word[32][32]={"do","if","while","include","main","define","for","char","int","float","long","double","switch","exit","break","continue","case","default","return","else","goto","printf","scanf","void","struct","union","enum","FILE"};

in=fopen("in.txt","r");out=fopen("out1.txt","w");

while(!feof(in)){

ch=fgetc(in);int i=0,j=0;flag=0;

if((ch=='+')||(ch=='-')||(ch=='*')||(ch=='/')||(ch=='<')||(ch=='>')||(ch=='=')||(ch=='%')||(ch=='?')) fprintf(out,"\n %c is operator \n",ch);

else if((ch=='#')||(ch=='~')||(ch==';')||(ch=='^')||(ch=='$')||(ch=='(')||(ch=='-')||(ch==')')||(ch=='{')||(ch=='}')||(ch=='[')||(ch==']')||(ch=='_')||(ch=='|')||(ch=='`')||(ch=='"')||(ch==':')||(ch=='.')||(ch==','))

fprintf(out,"\n %c specialsymbol \n",ch);

else if(isdigit(ch)){

str2[j]=ch;j++;ch=fgetc(in);while(isdigit(ch)){

str2[j]=ch;j++;ch=fgetc(in);

}str2[j]='\0';fprintf(out,"\n %s is Number\n",str2);//str2[j]='\0';

}else if(isalpha(ch)){

str1[i]=ch; i++; ch=fgetc(in);

while(isalnum(ch)){

str1[i]=ch; i++; ch=fgetc(in);

}str1[i]='\0';for( k=0;k<32;k++)if(strcmp(key_word[k],str1)==0)

flag=1;if(flag==1)

fprintf(out,"\n %s keyword\n",str1);else

fprintf(out,"\n %s identifier\n",str1);}

}fclose(in);fclose(out);

}

Execution:Cc scanner.c./a.out

In.txtvoid main(){int a = 10 , b = 20 , c ;c = a + b;printf("the sum is %d",c);}Out1.txtvoid keywordmain keyword) specialsymbol { specialsymbol int keyworda identifier= is operator 10 is Number, specialsymbol b identifier

= is operator 20 is Number, specialsymbol c identifier; specialsymbol c identifier= is operator a identifier+ is operator b identifierprintf keyword" specialsymbol the identifiersum identifieris identifier% is operator d identifier, specialsymbol c identifier; specialsymbol } specialsymbol

2. Write a program in lex to check if a no is octal, Hexadecimal etc.

Octal.l

%%[0][0-7]+ printf("%s\tIt is octal\n",yytext);[0][Xx][0-9A-Fa-f]+ printf("%s\tIt ishexadecimel\n",yytext);%%

Execution:

Lex octal.lCc lex.yy.c –ll./a.out

Output:0xA10xA1 It ishexadecimel

03210321 It is octal

3. Program in lex to capitalize all comments in a c program.

Capitalize.l

%{int c=0;%}comment [//]%%{comment} {c=1;}[A-Za-z] {if(c==1)printf("%s",caps(yytext));}

%%main(){yylex();}caps(char *str){int i=0;for(i=0;str[i]!='\0';i++)printf("%c",toupper(str[i]));}Execution: Output:lex cap1.l //mjcollegecc lex.yy.c –ll MJCOLLEGE./a.out

4. Program to implement a lex scanner.

Lscan.l

%{

#include <stdio.h>#include <string.h>void fn(int,int);int ln=1,n;

%}

d [0-9]n {d}+signo ([+/-]?{n})letter [A-Za-z]key "int"|"main"|"include"|"stdio.h"sc [;| |(|)|{|}|,|#|"|&|<|>]

%%

[\n] {ln++;}{key} {fn(ln,1);}[+|-|*|=] {fn(ln,2);}{letter} {fn(ln,3);}{sc} {fn(ln,4);}

%%

main(){

yyin=fopen("in.txt","r");yyout=fopen("out.txt","w");yylex();

}

void fn(int l,int n){

switch(n){

case 1: fprintf(yyout,"\n%d %s keyword",l,yytext); break;

case 2: fprintf(yyout,"\n%d %s operator",l,yytext);

break;

case 3: fprintf(yyout,"\n%d %s identifier",l,yytext); break;

default: fprintf(yyout,"\n%d %s special symbol",l,yytext); break;

}

}In.txt

#include<stdio.h>int main(){int a,b,c;c=a+b;}Out.txt1 # special symbol1 include keyword1 < special symbol1 stdio.h keyword1 > special symbol3 int keyword3 special symbol3 main keyword3 ( special symbol3 ) special symbol4 { special symbol5 int keyword5 special symbol5 a identifier5 , special symbol5 b identifier5 , special symbol5 c identifier5 ; special symbol6 c identifier6 = operator6 a identifier6 + operator6 b identifier6 ; special symbol7 } special symbol

5. Write a program to find first and follow elements.

First.c

#include<stdio.h>#include<ctype.h>

int main(){

int i,n,j,k; char str[10][10],f; printf("Enter the number of productions\n"); scanf("%d",&n); printf("Enter grammar\n"); for(i=0;i<n;i++) scanf("%s",&str[i]); for(i=0;i<n;i++) {

f= str[i][0]; int temp=i;

if(isupper(str[i][3])) {

repeat: for(k=0;k<n;k++) { if(str[k][0]==str[i][3]) { if(isupper(str[k][3])) { i=k; goto repeat; } else { printf("First(%c)=%c\n",f,str[k][3]); } } } } else { printf("First(%c)=%c\n",f,str[i][3]); }

i=temp; }

}

Output:

vi first.ccc first.c./a.out

Enter the number of productions5

Enter grammarS->ABA->abcB->(D)D->jklE->S

First(S)=aFirst(A)=aFirst(B)=(First(D)=jFirst(E)=a

Follow.c

#include<stdio.h>main(){

int np,i,j,k;char prods[10][10],follow[10][10],Imad[10][10];printf("enter no. of productions\n");scanf("%d",&np);printf("enter grammar\n");for(i=0;i<np;i++){

scanf("%s",&prods[i]);}

for(i=0; i<np; i++){

if(i==0) {

printf("Follow(%c) = $\n",prods[0][0]);//Rule1 }for(j=3;prods[i][j]!='\0';j++){

int temp2=j; //Rule-2: production A->xBb then everything in first(b) is in follow(B) if(prods[i][j] >= 'A' && prods[i][j] <= 'Z') {

if((strlen(prods[i])-1)==j) { printf("Follow(%c)=Follow(%c)\n",prods[i][j],prods[i][0]); } int temp=i; char f=prods[i][j]; if(!isupper(prods[i][j+1])&&(prods[i][j+1]!='\0')) printf("Follow(%c)=%c\n",f,prods[i][j+1]); if(isupper(prods[i][j+1])) {

repeat: for(k=0;k<np;k++) { if(prods[k][0]==prods[i][j+1]) { if(!isupper(prods[k][3])) { printf("Follow(%c)=%c\n",f,prods[k][3]); } else { i=k; j=2; goto repeat; } }

}}i=temp;

}j=temp2;

}}

}

Output :

vi follow.ccc follow.c./a.out

enter no. of productions4

Enter GrammarS->AB(C)B->DED->abcE->jkl

Follow(S) = $Follow(A)=aFollow(B)=(Follow(C)=)Follow(D)=jFollow(E)=Follow(B)

6. Write a Program to find canonical LR(0) collections. i.e. a). closure

closure.c

#include<stdio.h>#include<string.h>main(){ char A[10][10],items[20][20]={},closure[10][10]={},I[20][20],inputs[20][20]; int i,j,k,l=0,m=0,p=3,n=0,ilen[10],numi; int B[10][10]; char c,ch; int flag=0,numprods,temp,temp1,q;

printf("enter the number of productions\n"); scanf("%d",&numprods);

printf("Enter productions:\n");

for(i=0;i<=numprods;i++) { gets(A[i]); }

//finding items:

for(i=0;i<=numprods;i++) {

n = strlen(A[i]); for(j=0;j<(n-2);j++) { for(k=0;k<=(strlen(A[i]));k++) { if(p==k) { items[l][m] = '.'; m++; items[l][m] = A[i][k]; m++; } else { items[l][m++] = A[i][k]; }

}//k loop l++; p++; m = 0;

}//j loop p = 3;

}//i loop i = 0;

printf("Items are:\n");

while(l!=i) {

printf("%d)",i); puts(items[i]); printf("\n"); i++; }

printf("\n");

//Finding Closure:

strcpy(closure[0],items[0]); m=1;

j=0; do { if(closure[j][4]>='A' && closure[j][4] <= 'Z') { for(i=0;i<l;i++)// l gives the number of items { if((closure[j][4]==items[i][0])&&(items[i][4]=='.')) { strcpy(closure[++j],items[i-1]); m++; } } } else flag=1; }

while(flag==0); printf("Closure is:\n"); for(j=0;j<m;j++) puts(closure[j]); printf("\n");}

Output:

enter the number of productions2

Enter productions:A->CC->bItems are:0)A->.C

1)A->C.

2)C->.b

3)C->b.

Closure is:A->.CC->.b

7. Write a Program to construct SLR parsing table using program 6.

Slr.c

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

char a[8][5],b[7][5];int c[12][5];int w=0,e=0,x=0,y=0;int st2[12][2],st3[12];char sta[12],ch;

void v1(char,int);void v2(char,int,int,int);

int main() { int i,j,k,l=0,m=0,p=1,f=0,g,v=0,jj[12]; printf("\n\n\t*******Enter the Grammar Rules (max=3)*******\n\t"); for(i=0;i<3;i++) {

gets(a[i]);printf("\t");

} for(i=0;i<3;i++) {

for(j=0;j<strlen(a[i]);j++) {

for(k=0;k<strlen(a[i]);k++) {

if(p==k){

b[l][m]='.'; m+=1;

b[l][m]=a[i][k]; m+=1;

}else{

b[l][m]=a[i][k]; m++; }

}p++;l++;m=0;

} p=1;

} i=0; p=0; while(l!=i) { for(j=0;j<strlen(b[i]);j++) { if(b[i][j]=='.')

{ p++; }

} if(p==0) {

b[i][strlen(b[i])]='.';}

i++; p=0; } i=0; sleep(10);

printf("\n\t*******Your States will be*******\n\t"); while(l!=i) { printf("%d--> ",i); puts(b[i]); i++; printf("\t"); } printf("\n"); v1('A',l);

p=c[0][0]; m=0; while(m!=6) { for(i=0;i<st3[m];i++) { for(j=0;j<strlen(b[p]);j++) { if(b[p][j]=='.' && ((b[p][j+1]>=65 && b[p][j+1]<=90)||(b[p][j+1]>=97&&b[p][j+1]<=122)))

{ st2[x][0]=m; sta[x]=b[p][j+1]; v2(b[p][j+1],j,l,f); x++;

}else

{ if(b[p][j]=='.') {

st2[x][0]=m;sta[x]='S';st2[x][1]=m;x++;

}

}}

p=c[m][i+1]; } m++; p=c[m][0]; } g=0; p=0; m=0;

x=0;while(p!=11)

{ for(i=0;i<st3[p];i++) {

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

for(j=0;j<3;j++) { if(c[k][j]==c[p][j])

{ m++;

} }

if(m==3) { m=0; goto ac; } m=0;

} if(m!=3) { if(v==0) {

printf("\tI%d=",g); v++; jj[g]=p;

} printf("%d",c[p][i]);

}

}printf("\n");

g++; ac: p++; v=0; } sleep(10); printf("\t*******Your DFA will be *******");

for(i=0;i<9;i++) { printf("\n\t%d",st2[i][0]); printf("-->%c",sta[i]); }

getchar(); }

void v1(char ai,int kk){

int i,j; for(i=0;i<kk;i++) {

if(b[i][2]==ai&&b[i][1]=='.') { c[w][e]=i; e++; if(b[i][2]>=65 && b[i][2]<=90)

{ for(j=0;j<kk;j++) { if(b[j][0]==ai && b[j][1]=='.')

{ c[w][e]=j; e++; }

} }

}}st3[w]=e;w++;e=0;

}void v2(char ai,int ii,int kk,int tt)

{ int i,j,k; for(i=0;i<kk;i++)

{

if(b[i][ii]=='.'&& b[i][ii+1]==ai) { for(j=0;j<kk;j++)

{ if(b[j][ii+1]=='.' && b[j][ii]==ai) {

c[w][e]=j;e++;st2[tt][1]=j;if(b[j][ii+2]>=65 && b[j][ii+1]<=90)

{ for(k=0;k<kk;k++) { if(b[k][0]==b[j][ii+2] && b[k][1]=='.')

{ c[w][e]=k; e++; }

} }

} } if((b[i][ii+1]>=65 && b[i][ii+1]<=90) && tt==1) {

for(k=0;k<kk;k++) { if(b[k][0]==ai && b[k][1]=='.') {

c[w][e]=k; e++;

}}

}}

} st3[w]=e; w++; e=0; }

8. Write a program to implement a calculator using yacc. a). using lex b). without using lex

a)yacclex.l%{ #include "y.tab.h" extern int yylval;%}%%[0-9]+ {yylval=atoi(yytext); return NUM;}[\t]\n return 0;. return yytext[0];%%

yacclex.y:%{#include<stdio.h>#include<ctype.h>%}%token DIG%%cmd: E '\n' {printf("%d\n",$1);}

E:E'+'T {$$=$1+$3;} |T {$$=$1;} ;T:T'*'F {$$=$1*$3;} |F {$$=$1;} ;F:'('E')' {$$=$2;} |DIG {$$=$1;} ;%%yylex(){ int c; c=getchar(); if(isdigit(c)) { yylval=c-'0'; return DIG; } elsereturn c;}

Execution:lex lp6.l yacc lp6.y yacc -dyacc.y

(usage: yacc [-dlrtv] [-b file_prefix] [-o output_filename] [-p symbol_prefix] filename)

cc y.tab.c lex.yy.c -ly -ll ./a.out

output:5*420

b. yacc.y

%{ #include<stdio.h>%}%token DIG

%%cmd: E '\n' { printf("%d", $1);} ;E : E '+' T {$$=$1+$3;} | T {$$=$1;} ;

T : T '-' B {$$=$1-$3;} |B {$$=$1;} ;

B : B '/' Q{$$=$1/$3;} |Q {$$=$1;} ;

Q : Q '*' F {$$=$1*$3;} | F {$$=$1;} ;

F : '(' E ')' {$$=$2;} |DIG {$$=$1;} ;%%yylex(){int c;c=getchar();if(isdigit(c))

{

yylval=c-'0';return DIG;}

return c;}Execution:yacc yacc.ycc y.tab.c -ly./a.out

Output:5*315

9. Program to generate code.9.Gencode.c

#include<stdio.h>char stk[100],stktop=-1,cnt=0;

void push(char pchar){

stk[++stktop]=pchar;}

char pop(){

return stk[stktop--];}

char checkoperation(char char1){

char oper;if(char1=='+')

oper='A';else if(char1=='-')

oper='S';else if(char1== '*')

oper='M';else if(char1=='/')

oper='D';

return oper;}

int checknstore(char check){

int ret;if(check!='+' && check!='-' && check!='*' && check!='/' && check!='@'){

push(++cnt);if(stktop>0)printf("ST $%d\n",cnt);ret=1;

}

elseret=0;

return ret;}

void main()

{char msg[100],op1,op2,operation;int i,val;while(scanf("%s",msg)!=EOF){

cnt=0;stktop=-1;for(i=0;msg[i]!='\0';i++){

if((msg[i]>='A' && msg[i]<='Z')|| (msg[i]>='a' && msg[i]<='z'))push(msg[i]);

else{

op1=pop();op2=pop();printf("L %c\n",op2);operation= checkoperation(msg[i]);printf("%c %c\n",operation,op1);val=checknstore(msg[i+1]);while(val==0){

op1=pop();cnt--;operation=checkoperation(msg[++i]);if(operation=='S'&&stktop>=-1){printf("N\n");operation='A';}printf("%c %s\n", operation,op1);val=checknstore(msg[i+1]);

}}

}}

}

Output:AB+C*D+L AA BL ?M CL ? A D

10. Program to optimize the code.

Codeopti.c

/* Code Optimization prog*/

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

main(){

char str[25][50], forLoopParam[90], rightHandParam[10][40], leftHandParam[90];int j=0,k=0,i=0,m=0,n=0,q=0,s=0;int flag[10]={0},count[10]={0};printf("\n Input the loop to be optimized:\n");

/* PROCESSING FIRST LINE -- FOR DECLARATION*/

gets(str[0]);while(str[k][i++]!=';');while(str[k][i++]!=';');

while(str[k][i]!=')'){

if(isalpha(str[k][i])) forLoopParam[j++]=str[k][i];

i++;}i=0;

/* J IS INDEX TO CHANGEARRAY*/

/* PROCESSING SECOND LINE --{*/

puts(str[0]);gets(str[0]);

while(str[0][i++]!='{');puts(str[0]);k=0;while(gets(str[k]) && str[k][0]!='}'){

while(str[k][i++]!='=');leftHandParam[n++]=str[k][i-2];k++;i=0;

}

/*N IS INDEX TO TEMPCHANGE ARRAY*//* TEMPCHANGE ARRAY STORES LEFT SIDE PARAMETERS OF ASSIGNMENT OPERATIONS */

for(m=0,i=0;m<k;m++){

while(str[m][i++]!='=');while(str[m][i]!=';'){

if(isalpha(str[m][i]))rightHandParam[m][count[m]++]=str[m][i];i++;

} i=0;}

/* Q IS INDEX TO TEMP2*//* TEMP2 STORES RIGHT HAND SIDE PARAMETERS OF STATEMENTS*//*COMPARING LEFT-HAND PARAMETERS WITH RIGHT-HAND PARAMETERS*/

for(m=0;m<k;m++) for(s=0;s<count[m];s++) for(i=0;i<n;i++)

{ if(rightHandParam[m][s]==leftHandParam[i])

flag[m]=1;}

//COMPARING LEFT-HAND SIDE PARAMETERS WITH FOR-LOOP PARAMETERS

for(i=0;i<k;i++) for(q=0;q<j;q++) {

if(leftHandParam[i]==forLoopParam[q])flag[i]=1;

}

//COMPARING RIGHT-HAND PARAMETERS WITH FOR-LOOP PARAMETERS

for(m=1;m<k;m++) for(s=0;s<count[m];s++) for(i=0;i<j;i++)

{ if(rightHandParam[m][s]==forLoopParam[i])

flag[m]=1;}

//DISPLAYING STATEMENTS OF FOR LOOP WHICH CANT BE OPTIMIZED IN THE FOR LOOP

for(i=0;i<k;i++) if(flag[i]==1) puts(str[i]);

//DISPLAYING STATEMENTS OF FOR LOOP WHICH CANT BE OPTIMIZED OUTSIDE FOR LOOP

puts(str[k]); //DISPLAYING THE END-FLOWER BRACE

for(i=0;i<k;i++) if(flag[i]==0)

printf("\t\t %s \t\t",str[i]);}

Execution:Cc codeopti.c./a.out

Output:Input the loop to be optimized:for(i=0;i<9;i++)for(i=0;i<9;i++){{c=n+m;i=i+j;}i=i+j;} c=n+m;

11.Program to implement recursive descent parser

rec1parse.c

#include<stdio.h>#include<stdlib.h>char token;int exp(void);int term(void);int fac(void);void match(char etoken)

{ if (token==etoken) token=getchar(); else

{printf("ERROR\n");exit(1);} } main() { token=getchar(); exp(); if (token=='\n') printf("SUCCESS\n"); else printf("ERROR\n"); } int exp(void) {term(); while((token=='+') ||(token=='-')) switch(token) { case '+': match('+'); term(); break;

case '-': match('-'); term(); break; } } int term(void) {fac(); while(token=='*') { match('*'); fac(); } } int fac(void) { if (token=='(') { match('('); exp(); match(')'); } else if(isdigit(token)) { match(token); } else {printf("ERROR\n");exit(1);} }

Output :1+9+8

vi rec1parse.c SUCCESS cc rec1parse.c 8+7*./a.out ERROR