%{
#include<stdio.h>
#include<stdlib.h>
int regs[30];
%}
%token NUMBER LETTER
%left PLUS MINUS
%left MULT DIV
%%
prog: prog st | ; //when I remove this line the error goes
st : E {printf("ans %d", $1);}| LETTER '=' E {regs[$1] = $3; printf("variable contains %d",regs[$1]);};
E : E PLUS E{$$ = $1 + $3;} //addition
| E MINUS E{$$ = $1 - $3 ;} //subtraction
| MINUS E{$$ = -$2;}
| E MULT E{$$ = $1 * $3 ;}
| E DIV E { if($3)$$= $1 / $3; else yyerror("Divide by 0");}
/*|LBRACE E RBRACE{$$= $2;}
| RBRACE E LBRACE{yyerror("Wrong expression");} */
| NUMBER {$$ = $1;}
| LETTER {$$ = regs[$1];}
;
%%
int main(void)
{
printf("Enter Expression: ");
yyparse();
return 0;
}
int yyerror(char *msg)
{
printf("%s", msg);// printing error
exit(0);
}
I am not able to resolve the conflicts. Also I am getting a segmentation fault when I run it with some edits. I am using yacc and lex for the same.
The two shift-reduce conflicts are the result of the fact that you don't require any explicit separator between statements. Because of that, a = b - 3 could be interpreted as one statement or as two (a = b; - 3). The second interpretation may not seem very natural to you but it is easily derived by the grammar.
In addition, your use of unary minus leads to an incorrect parse of -2/3 as -(2/3) instead of (-2)/3. (You may or may not find this serious, since it has few semantic consequences with these particular operators.) This particular issue and a correct resolution is discussed in the bison manual, and in many many other internet resources.
Both of these explanations are made a bit more visible if you use the -v command line option to bison to produce a description of the parser. See Understanding your parser (again, in the bison manual).