Write a function Add() that returns sum of two integers. The function should not use any of the arithmetic operators (+, ++, –, -, .. etc).
Method 1:
C++
#include <iostream>using namespace std;int add(int a, int b){ // for loop will start from 1 and move till the value of // second number , first number(a) is incremented in for // loop for (int i = 1; i <= b; i++) a++; return a;}int main(){ // first number is 10 and second number is 32 , for loop // will start from 1 and move till 32 and the value of a // is incremented 32 times which will give us the total // sum of two numbers int a = add(10, 32); cout << a; return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
#include <stdio.h>int add(int a, int b){ // for loop will start from 1 and move till the value of // second number , first number(a) is incremented in for // loop for (int i = 1; i <= b; i++) a++; return a;}int main(){ // first number is 10 and second number is 32 , for loop // will start from 1 and move till 32 and the value of a // is incremented 32 times which will give us the total // sum of two numbers int a = add(10, 32); printf("%d", a); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
import java.util.*;class GFG { static int add(int a, int b) { // for loop will start from 1 and move till the // value of second number , first number(a) is // incremented in for loop for (int i = 1; i <= b; i++) a++; return a; } public static void main(String[] args) { // first number is 10 and second number is 32 , for // loop will start from 1 and move till 32 and the // value of a is incremented 32 times which will // give us the total sum of two numbers int a = add(10, 32); System.out.print(a); }}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Python implementationdef add(a, b): # for loop will start from 1 and move till the value of second number , # first number(a) is incremented in for loop for i in range(1, b + 1): a = a + 1 return a# driver code# first number is 10 and second number is 32 , for loop# will start from 1 and move till 32 and the value of a# is incremented 32 times which will give us the total# sum of two numbersa = add(10, 32)print(a)# This code is contributed by Aditya Kumar (adityakumar129) |
C#
using System;public class GFG { static int add(int a, int b) { for (int i = 1; i <= b; i++) // for loop will start from 1 and move till the value of second // number , first number(a) is incremented in for loop { a++; } return a; } public static void Main(String[] args) { int a = add(10, 32); // first number is 10 and second number is 32 , for loop will start Console.Write(a); // from 1 and move till 32 and the value of a is incremented 32 times // which will give us the total sum of two numbers }}// This code is contributed by Rajput-Ji |
Javascript
<script> function add(a , b) { // for loop will start from 1 and move till the value of second // number , first number(a) is incremented in for loop for (i = 1; i <= b; i++) { a++; } return a; } // first number is 10 and second number is 32 , for loop will start var a = add(10, 32); // from 1 and move till 32 and the value of a is incremented 32 times // which will give us the total sum of two numbers document.write(a); // This code is contributed by Rajput-Ji</script> |
Output
42
Time Complexity: O(b)
Auxiliary Space: O(1)
Above is simple Half Adder logic that can be used to add 2 single bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result.
C++
// C++ Program to add two numbers // without using arithmetic operator #include <bits/stdc++.h>using namespace std;int Add(int x, int y) { // Iterate till there is no carry while (y != 0) { // carry should be unsigned to // deal with -ve numbers // carry now contains common //set bits of x and y unsigned carry = x & y; // Sum of bits of x and y where at //least one of the bits is not set x = x ^ y; // Carry is shifted by one so that adding // it to x gives the required sum y = carry << 1; } return x; } // Driver codeint main() { cout << Add(10, 32); return 0; } // This code is contributed by rathbhupendra |
C
// C Program to add two numbers// without using arithmetic operator#include<stdio.h>int Add(int x, int y){ // Iterate till there is no carry while (y != 0) { // carry now contains common //set bits of x and y unsigned carry = x & y; // Sum of bits of x and y where at //least one of the bits is not set x = x ^ y; // Carry is shifted by one so that adding // it to x gives the required sum y = carry << 1; } return x;}int main(){ printf("%d", Add(15, 32)); return 0;} |
Java
// Java Program to add two numbers// without using arithmetic operatorimport java.io.*;class GFG { static int Add(int x, int y) { // Iterate till there is no carry while (y != 0) { // carry now contains common // set bits of x and y int carry = x & y; // Sum of bits of x and // y where at least one // of the bits is not set x = x ^ y; // Carry is shifted by // one so that adding it // to x gives the required sum y = carry << 1; } return x; } // Driver code public static void main(String arg[]) { System.out.println(Add(15, 32)); }}// This code is contributed by Anant Agarwal. |
Python3
# Python3 Program to add two numbers# without using arithmetic operatordef Add(x, y): # Iterate till there is no carry while (y != 0): # carry now contains common # set bits of x and y carry = x & y # Sum of bits of x and y where at # least one of the bits is not set x = x ^ y # Carry is shifted by one so that # adding it to x gives the required sum y = carry << 1 return xprint(Add(15, 32))# This code is contributed by# Smitha Dinesh Semwal |
C#
// C# Program to add two numbers// without using arithmetic operatorusing System;class GFG { static int Add(int x, int y) { // Iterate till there is no carry while (y != 0) { // carry now contains common // set bits of x and y int carry = x & y; // Sum of bits of x and // y where at least one // of the bits is not set x = x ^ y; // Carry is shifted by // one so that adding it // to x gives the required sum y = carry << 1; } return x; } // Driver code public static void Main() { Console.WriteLine(Add(15, 32)); }}// This code is contributed by vt_m. |
Javascript
<script>// Javascript Program to add two numbers // without using arithmetic operator function Add(x, y) { // Iterate till there is no carry while (y != 0) { // carry now contains common //set bits of x and y let carry = x & y; // Sum of bits of x and y where at //least one of the bits is not set x = x ^ y; // Carry is shifted by one so that adding // it to x gives the required sum y = carry << 1; } return x; } //driver code document.write(Add(15, 32)); // This code is contributed by Surbhi Tyagi </script> |
PHP
<?php// PHP Program to add two numbers// without using arithmetic operatorfunction Add( $x, $y){ // Iterate till there is // no carry while ($y != 0) { // carry now contains common //set bits of x and y $carry = $x & $y; // Sum of bits of x and y where at //least one of the bits is not set $x = $x ^ $y; // Carry is shifted by one // so that adding it to x // gives the required sum $y = $carry << 1; } return $x;} // Driver Code echo Add(15, 32);// This code is contributed by anuj_67.?> |
Output
42
Time Complexity: O(log y)
Auxiliary Space: O(1)
Following is the recursive implementation for the same approach.
C++
int Add(int x, int y){ if (y == 0) return x; else return Add( x ^ y,(unsigned) (x & y) << 1);}// This code is contributed by shubhamsingh10 |
C
int Add(int x, int y){ if (y == 0) return x; else return Add( x ^ y, (unsigned)(x & y) << 1);} |
Java
static int Add(int x, int y){ if (y == 0) return x; else return Add(x ^ y, (x & y) << 1);}// This code is contributed by subham348 |
Python3
def Add(x, y): if (y == 0): return x else: return Add( x ^ y, (x & y) << 1) # This code is contributed by subhammahato348 |
C#
static int Add(int x, int y){ if (y == 0) return x; else return Add(x ^ y, (x & y) << 1);}// This code is contributed by subhammahato348 |
Javascript
function Add(x, y){ if (y == 0) return x; else return Add(x ^ y, (x & y) << 1);}// This code is contributed by Ankita saini |
Time Complexity: O(log k), where k=x&y
Auxiliary Space: O(log k).
Another Method:
This method also add 2 numbers without using any arithmetic operators.
Steps:
Follow the given steps to solve the problem:
- First calculate 2 to the power given numbers.
- Then just use log (base 2) which returns their sum.
- Finally return the ans.
Below is the Implementation of the above approach:
C++
// CPP code of the above approach#include <bits/stdc++.h>using namespace std;// Function to add 2 numbers using log and powerint Add(int a, int b){ int ans = log2(pow(2, a) * pow(2, b)); return ans;}int main(){ int a = 10, b = 5; cout << Add(a, b) << endl; return 0;}// This code is contributed by Susobhan Akhuli |
Java
// Java code of the above approachimport java.lang.Math;public class GFG { // Function to add 2 numbers using log and power static int add(int a, int b) { int ans = (int)(Math.log(Math.pow(2, a) * Math.pow(2, b)) / Math.log(2)); return ans; } public static void main(String[] args) { int a = 10, b = 5; System.out.println(add(a, b)); }}// This code is contributed by Susobhan Akhuli |
Output
15
Time Complexity: O(log N), for pow function.
Auxiliary Space: O(1).
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.
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