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infix to postfix

Azad Mahmud answered on July 22, 2023 Popularity 10/10 Helpfulness 2/10

answer infix to postfix

related Convert Infix Expression to Postfix

infix to postfix

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// 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 whatever

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Contributed on Jul 22 2023

Azad Mahmud

0 Answers Avg Quality 2/10

Closely Related Answers

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 e

Link to this answer
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Contributed on Sep 10 2023

Ashamed Angelfish

0 Answers Avg Quality 2/10

infix to postfix

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