Yacc Program to evaluate a given arithmetic expression

Prerequisite – Introduction to YACC
Problem: Write a YACC program to evaluate a given arithmetic expression consisting of ‘+’, ‘-‘, ‘*’, ‘/’ including brackets.
Examples:

Input: 7*(5-3)/2
Output: 7

Input: 6/((3-2)*(-5+2))
Output: -2 

Lexical Analyzer Source Code:

C

%{
    /* Definition section*/
    #include "y.tab.h"
    extern yylval;
%}

%%
[0-9]+    {
              yylval = atoi(yytext);
              return NUMBER;
            }

[a-zA-Z]+    { return ID; }
[ \t]+         ;  /*For skipping whitespaces*/

\n            { return 0; }
.            { return yytext[0]; }

%%

Parser Source Code:

C

%{
    /* Definition section */
  #include <stdio.h>
%}

%token NUMBER ID
// setting the precedence
// and associativity of operators
%left '+' '-'
%left '*' '/'

/* Rule Section */
%%
E : T        {
                printf("Result = %d\n", $$);
                return 0;
            }

T :
    T '+' T { $$ = $1 + $3; }
    | T '-' T { $$ = $1 - $3; }
    | T '*' T { $$ = $1 * $3; }
    | T '/' T { $$ = $1 / $3; }
    | '-' NUMBER { $$ = -$2; }
    | '-' ID { $$ = -$2; }
    | '(' T ')' { $$ = $2; }
    | NUMBER { $$ = $1; }
    | ID { $$ = $1; };
% %

int main() {
    printf("Enter the expression\n");
    yyparse();
}

/* For printing error messages */
int yyerror(char* s) {
    printf("\nExpression is invalid\n");
}

Output:

Notes:
Yacc programs are generally written in 2 files one for lex with .l extension(for tokenization and send the tokens to yacc) and another for yacc with .y extension (for grammar evaluation and result evaluation).
Steps for execution of Yacc program:

yacc -d sample_yacc_program.y
lex sample_lex_program.l
cc lex.yy.c y.tab.c -ll
./a.out