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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