infix to postfix expression converter in "C" language

 program:

#include<stdio.h>

void infix_to_postfix(char[],char[]);

int precedence(char);

void main()

{

char infix[20],postfix[20];

printf("enter infix string : ");

scanf("%s",infix);

infix_to_postfix(infix,postfix);

printf("\npostfix exp= %s",postfix);

}

void infix_to_postfix(char infix[],char postfix[])

{

char s[20],symbol;

int top=-1,i=0,j=0;

while(infix[i]!='\0')

{

symbol=infix[i];

switch(symbol)

{

case '(':++top;

         s[top]=symbol;

         break;

    case ')':while(s[top]!='(')

             {

              postfix[j++]=s[top];

              top--;

}

          top--;

          break;

    case '+':

    case '-':

    case '*':

    case '^':

    case '/':

          while(precedence(symbol)<=precedence(s[top]) && top!=-1)

                   postfix[j++]=s[top--];

           s[++top]=symbol;

           break;

    default:postfix[j++]=symbol;       //to add operand to postfix exp

}

i++;

}

while(top!=-1)

{

postfix[j++]=s[top];

top--;

}

postfix[j]='\0';

}

int precedence(char x)

{

switch(x)

{

case '+':

case '-':return(1);

case '*':

case '^':return(3);

case '%':

case '/':return(2);

default:return(0);

}

}



output:

enter infix expression : a+b*c

postfix exp = abc*+

that's it folks.
comment what you need and happy coding.

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