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YACC program to implement a Calculator and recognize a valid Arithmetic expression
  • Last Updated : 26 Aug, 2020

Problem: YACC program to implement a Calculator and recognize a valid Arithmetic expression.

Explanation:
Yacc (for “yet another compiler compiler.”) is the standard parser generator for the Unix operating system. An open source program, yacc generates code for the parser in the C programming language. The acronym is usually rendered in lowercase but is occasionally seen as YACC or Yacc.

Examples:

Input: 4+5 
Output: Result=9
Entered arithmetic expression is Valid

Input: 10-5
Output: Result=5
Entered arithmetic expression is Valid

Input: 10+5-
Output: 
Entered arithmetic expression is Invalid

Input: 10/5
Output: Result=2
Entered arithmetic expression is Valid

Input: (2+5)*3
Output: Result=21
Entered arithmetic expression is Valid

Input: (2*4)+
Output: 
Entered arithmetic expression is Invalid

Input: 2%5
Output: Result=2
Entered arithmetic expression is Valid 

Lexical Analyzer Source Code:




%{
   /* Definition section */
  #include<stdio.h>
  #include "y.tab.h"
  extern int yylval;
%}
  
/* Rule Section */
%%
[0-9]+ {
          yylval=atoi(yytext);
          return NUMBER;
  
       }
[\t] ;
  
[\n] return 0;
  
. return yytext[0];
  
%%
  
int yywrap()
{
 return 1;
}

Parser Source Code :




%{
   /* Definition section */
  #include<stdio.h>
  int flag=0;
%}
  
%token NUMBER
  
%left '+' '-'
  
%left '*' '/' '%'
  
%left '(' ')'
  
/* Rule Section */
%%
  
ArithmeticExpression: E{
  
         printf("\nResult=%d\n", $$);
  
         return 0;
  
        };
 E:E'+'E {$$=$1+$3;}
  
 |E'-'E {$$=$1-$3;}
  
 |E'*'E {$$=$1*$3;}
  
 |E'/'E {$$=$1/$3;}
  
 |E'%'E {$$=$1%$3;}
  
 |'('E')' {$$=$2;}
  
 | NUMBER {$$=$1;}
  
 ;
  
%%
  
//driver code
void main()
{
   printf("\nEnter Any Arithmetic Expression which 
                   can have operations Addition, 
                   Subtraction, Multiplication, Division, 
                          Modulus and Round brackets:\n");
  
   yyparse();
   if(flag==0)
   printf("\nEntered arithmetic expression is Valid\n\n");
}
  
void yyerror()
{
   printf("\nEntered arithmetic expression is Invalid\n\n");
   flag=1;
}

Output:




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