Sunday 14 August 2016

Lab Program 4 Infix to Postfix 15CSL38 Data Structures in C Lab

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)

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