Lab Program 4:
Design, Develop and Implement 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("\nEnter the valid infix
expression:\t");
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*e
The entered
infix expression is :
(a+b)+c/d*e
The
corresponding postfix expression is :
ab+cd/e*+
Also Credits to:
Manoj Taleka (manoj89biet@gmail.com)
Yashaswini Jogi (jogi.yash@gmail.com)
