Add two numbers without using arithmetic operators

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Given two integers a and b, the task is to find the sum of a and b without using + or - operators.

Examples: 

Input: a = 10, b = 30
Output: 40

Input: a = -1, b = 2
Output: 1

Approach:

The approach is to add two numbers using bitwise operations. Let's first go through some observations:

• a & b will have only those bits set which are set in both a and b.
• a ^ b will have only those bits set which are set in either a or b but not in both.

If we want to calculate the sum of a and b such that a and b has no common set bit, then a ^ b is same as a + b. So, we can say that a + b without carry = a ^ b.

How to handle the case when we have carry, that is a and b has common set bits?

To calculate the carry, we know that carry will only have the common set bits of a and b, shifted 1 place to the left. So, we can say that carry = (a & b) << 1.

Now, the problem is reduced to calculating the sum of (a + b without carry) and carry. This is again the same problem: sum of two numbers where the first number = (a + b without carry) and second number = carry. So, we can repeatedly calculate carry and sum without carry, till carry is not reduced to 0.

Working:

Note: The above approach will work perfectly for languages like C++, Java, C# and JavaScript in which an integer has a size of 32 bits but in Python, we need to handle the overflow separately by creating a 32-bit mask of 1's.

Below is the implementation of the above algorithm:

// C++ Program to add two numbers without using arithmetic operators 

#include <bits/stdc++.h>
using namespace std;

// function to add two numbers without using arithmetic operators
int sum(int a, int b) {
    
    // Iterate till there is no carry 
    while (b != 0) { 
        
        // carry contains common set bits of a and b, left shifted by 1
        int carry = (a & b) << 1;

        // Update a with (a + b without carry)
        a = a ^ b;
      
        // Update b with carry
        b = carry; 
    } 
    return a;
} 

int main() {
    int a = -1, b = 2;
  
    cout << sum(a, b);
    return 0; 
} 

// C Program to add two numbers without using arithmetic operators 

#include <stdio.h>

// function to add two numbers without using arithmetic operators
int sum(int a, int b) {
    
    // 32 bit mask in hexadecimal
    long mask = 0xffffffff;

    // Iterate till there is no carry 
    while ((b & mask) != 0) { 
        
        // carry contains common set bits of a and b, left shifted by 1
        int carry = ((a & b) & mask) << 1;

        // Update a with (a + b without carry)
        a = a ^ b;
      
        // Update b with carry
        b = carry; 
    } 
    return a & mask;
} 

int main() {
    int a = -1, b = 2;
  
    printf("%d\n", sum(a, b));
    return 0; 
}

// Java Program to add two numbers without using arithmetic operators 

class GfG {

    // function to add two numbers without using arithmetic operators
    static int sum(int a, int b) {
        
        // Iterate till there is no carry 
        while (b != 0) { 
            
            // carry contains common set bits of a and b, left shifted by 1
            int carry = (a & b) << 1;

            // Update a with (a + b without carry)
            a = a ^ b;
          
            // Update b with carry
            b = carry; 
        } 
        return a;
    } 

    public static void main(String[] args) {
        int a = -1, b = 2;
      
        System.out.println(sum(a, b));
    } 
}

# Python Program to add two numbers without using arithmetic operators 

# function to add two numbers without using arithmetic operators
def sum(a, b):
    
    # 32 bit mask in hexadecimal
    mask = 0xffffffff

    # Iterate till there is no carry 
    while (b & mask) != 0:
        
        # carry contains common set bits of a and b, left shifted by 1
        carry = (a & b) << 1

        # Update a with (a + b without carry)
        a = a ^ b
      
        # Update b with carry
        b = carry 
    
    return a & mask if b > 0 else a

a = -1
b = 2

print(sum(a, b))

// C# Program to add two numbers without using arithmetic operators 

using System;

class GfG {
    
    // function to add two numbers without using arithmetic operators
    static int Sum(int a, int b) {
        
        // Iterate till there is no carry 
        while (b != 0) {
            
            // carry contains common set bits of a and b, left shifted by 1
            int carry = (a & b) << 1;

            // Update a with (a + b without carry)
            a = a ^ b;

            // Update b with carry
            b = carry;
        }
        return a;
    }

    static void Main() {
        int a = -1, b = 2;

        Console.WriteLine(Sum(a, b));
    }
}

// JavaScript Program to add two numbers without using arithmetic operators 

// function to add two numbers without using arithmetic operators
function sum(a, b) {
    
    // Iterate till there is no carry 
    while (b !== 0) {
        
        // carry contains common set bits of a and b, left shifted by 1
        let carry = (a & b) << 1;

        // Update a with (a + b without carry)
        a = a ^ b;
      
        // Update b with carry
        b = carry; 
    } 
    return a;
}

const a = -1, b = 2;

console.log(sum(a, b));

1

Time Complexity: O(max(number of bits in a, number of bits in b))
Auxiliary Space: O(1)