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:
%{
#include<stdio.h>
#include "y.tab.h"
extern int yylval;
%}
%%
[0-9]+ {
yylval= atoi (yytext);
return NUMBER;
}
[\t] ;
[\n] return 0;
. return yytext[0];
%%
int yywrap()
{
return 1;
}
|
Parser Source Code :
%{
#include<stdio.h>
int flag=0;
%}
%token NUMBER
%left '+' '-'
%left '*' '/' '%'
%left '(' ')'
%%
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;}
;
%%
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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