Lex and Yacc program for grammar evaluation

 ####### pr_3.l ##########
%{
#include<stdio.h>
#include "y.tab.h"
extern int yylval;
%}

%%
[0-9]+ {yylval=atoi(yytext);return(id);}
[\n] {return(0);}
. {return(yytext[0]);}
%%

######## pr_3.y #########

%{

#include<stdio.h>

int cnt=0;

%}

%token id

%left '+' '-'

%left '*' '/'

%%

edash : E {printf("\n\t\tAnswer=%d\n",$$);}

;

E : E'+'E {$$=$1+$3;printf("\n\t%d)E->E+E Rule is reduced ---- %d -> %d + %d",cnt++,$$,$1,$3);}

|

E'-'E {$$=$1-$3;printf("\n\t%d)E->E-E Rule is reduced ---- %d -> %d - %d",cnt++,$$,$1,$3);}

|

E'*'E {$$=$1*$3;printf("\n\t%d)E->E*E Rule is reduced ---- %d -> %d * %d",cnt++,$$,$1,$3);}

|

E'/'E {$$=$1/$3;printf("\n\t%d)E->E/E Rule is reduced ---- %d -> %d / %d",cnt++,$$,$1,$3);}

|

id

;

%%

int main()

{

printf("\n** We will see the execution sequence **");

printf("\n\tEnter Input = ");

yyparse();

return(0);

}

void yyerror (char *s)

{

fprintf (stderr,"%s", s);

}

/* OP

** We will see the execution sequence **

Enter Input = 10+5*2-1/1

0)E->E/E Rule is reduced ---- 1 -> 1 / 1

1)E->E-E Rule is reduced ---- 1 -> 2 - 1

2)E->E*E Rule is reduced ---- 5 -> 5 * 1

3)E->E+E Rule is reduced ---- 15 -> 10 + 5

Answer=15

pr@prashant-HP:~/LEX&YACC$ 

*/