4. Develop a Program in C for converting an Infix Expression to Postfix Expression. Program should support for both parenthesized and free parenthesized expressions with the operators: +, -, *, /, (Remainder), ^ (Power) and alphanumeric operands.
#include<stdio.h>
#include<stdlib.h>
void evaluate();
void push(char);
char pop();
int prec(char);
char infix[30], postfix[30], stack[30];
int top = -1;
void main()
{
printf("\n Enter the valid infix expression:");
scanf("%s", infix);
evaluate();
printf("\nThe entered infix expression is :\n %s \n", infix);
printf("\nThe corresponding postfix expression is :\n %s \n", postfix);
}
void evaluate()
{
int i = 0, j = 0;
char symb, temp;
push('#');
for (i = 0; infix[i] != '\0'; i++)
{
symb = infix[i];
switch (symb)
{
case '(':
push(symb);
break;
case ')':
temp = pop();
while (temp != '(')
{
postfix[j] = temp;
j++;
temp = pop();
}
break;
case '+':
case '-':
case '*':
case '/':
case '%':
case '^':
case '$':
while (prec(stack[top]) >= prec(symb))
{
temp = pop();
postfix[j] = temp;
j++;
}
push(symb);
break;
default:
postfix[j] = symb;
j++;
}
}
while (top > 0)
{
temp = pop();
postfix[j] = temp;
j++;
}
postfix[j] = '\0';
}
void push(char item)
{
top = top + 1;
stack[top] = item;
}
char pop()
{
char item;
item = stack[top];
top = top - 1;
return item;
}
int prec(char symb)
{
int p;
switch (symb)
{
case '#':
p = -1;
break;
case '(':
case ')':
p = 0;
break;
case '+':
case '-':
p = 1;
break;
case '*':
case '/':
case '%':
p = 2;
break;
case '^':
case '$':
p = 3;
break;
}
return p;
}OUTPUT:
Enter the valid infix expression:(a+b)*c/d^5%1
The entered infix expression is :
(a+b)*c/d^5%1
The corresponding postfix expression is :
ab+c*d5^/1%