Breaking News: Grepper is joining You.com. Read the official announcement!

Convert Infix Expression to Postfix

Ashamed Angelfish answered on July 22, 2023 Popularity 9/10 Helpfulness 1/10

answer Convert Infix Expression to Postfix

Convert Infix Expression to Postfix

Comment

// C code to convert infix to postfix expression
  
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
  
#define MAX_EXPR_SIZE 100
  
// Function to return precedence of operators
int precedence(char operator)
{
    switch (operator) {
    case '+':
    case '-':
        return 1;
    case '*':
    case '/':
        return 2;
    case '^':
        return 3;
    default:
        return -1;
    }
}
  
// Function to check if the scanned character 
// is an operator
int isOperator(char ch)
{
    return (ch == '+' || ch == '-' || ch == '*' || ch == '/'
            || ch == '^');
}
  
// Main functio to convert infix expression
// to postfix expression
char* infixToPostfix(char* infix)
{
    int i, j;
    int len = strlen(infix);
    char* postfix = (char*)malloc(sizeof(char) * (len + 2));
    char stack[MAX_EXPR_SIZE];
    int top = -1;
  
    for (i = 0, j = 0; i < len; i++) {
        if (infix[i] == ' ' || infix[i] == '\t')
            continue;
        
        // If the scanned character is operand
        // add it to the postfix expression
        if (isalnum(infix[i])) {
            postfix[j++] = infix[i];
        }
        
        // if the scanned character is '('
        // push it in the stack
        else if (infix[i] == '(') {
            stack[++top] = infix[i];
        }
        
        // if the scanned character is ')'
        // pop the stack and add it to the 
        // output string until empty or '(' found
        else if (infix[i] == ')') {
            while (top > -1 && stack[top] != '(')
                postfix[j++] = stack[top--];
            if (top > -1 && stack[top] != '(')
                return "Invalid Expression";
            else
                top--;
        }
        
        // If the scanned character is an operator
        // push it in the stack
        else if (isOperator(infix[i])) {
            while (top > -1
                   && precedence(stack[top])
                          >= precedence(infix[i]))
                postfix[j++] = stack[top--];
            stack[++top] = infix[i];
        }
    }
  
    // Pop all remaining elements from the stack
    while (top > -1) {
        if (stack[top] == '(') {
            return "Invalid Expression";
        }
        postfix[j++] = stack[top--];
    }
    postfix[j] = '\0';
    return postfix;
}
  
// Driver code
int main()
{
    char infix[MAX_EXPR_SIZE] = "a+b*(c^d-e)^(f+g*h)-i";
    
    // Function call
    char* postfix = infixToPostfix(infix);
    printf("%s\n", postfix);
    free(postfix);
    return 0;
}

Popularity 9/10 Helpfulness 1/10 Language whatever

Source: www.geeksforgeeks.org

Tags: expression postfix whatever

Link to this answer
Share Copy Link

Contributed on Sep 10 2023

Ashamed Angelfish

0 Answers Avg Quality 2/10

Closely Related Answers

infix to postfix

Comment

// C++ code to convert infix expression to postfix
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to return precedence of operators
int prec(char c)
{
    if (c == '^')
        return 3;
    else if (c == '/' || c == '*')
        return 2;
    else if (c == '+' || c == '-')
        return 1;
    else
        return -1;
}
  
// The main function to convert infix expression
// to postfix expression
void infixToPostfix(string s)
{
  
    stack<char> st;
    string result;
  
    for (int i = 0; i < s.length(); i++) {
        char c = s[i];
  
        // If the scanned character is
        // an operand, add it to output string.
        if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')
            || (c >= '0' && c <= '9'))
            result += c;
  
        // If the scanned character is an
        // ‘(‘, push it to the stack.
        else if (c == '(')
            st.push('(');
  
        // If the scanned character is an ‘)’,
        // pop and add to output string from the stack
        // until an ‘(‘ is encountered.
        else if (c == ')') {
            while (st.top() != '(') {
                result += st.top();
                st.pop();
            }
            st.pop();
        }
  
        // If an operator is scanned
        else {
            while (!st.empty()
                   && prec(s[i]) <= prec(st.top())) {
                result += st.top();
                st.pop();
            }
            st.push(c);
        }
    }
  
    // Pop all the remaining elements from the stack
    while (!st.empty()) {
        result += st.top();
        st.pop();
    }
  
    cout << result << endl;
}
  
// Driver code
int main()
{
    string exp = "a+b*(c^d-e)^(f+g*h)-i";
  
    // Function call
    infixToPostfix(exp);
    
    return 0;
}

Popularity 10/10 Helpfulness 2/10 Language whatever

Source: www.geeksforgeeks.org

Tags: postfix

Link to this answer
Share Copy Link

Contributed on Jul 22 2023

Azad Mahmud

0 Answers Avg Quality 2/10

Convert Infix Expression to Postfix

Contents

More Related Answers

Convert Infix Expression to Postfix

Closely Related Answers

infix to postfix

Grepper

Documentation

Social

Legal

Contact

Oops, You will need to install Grepper and log-in to perform this action.