# Bitwise recursive addition of two integers

• Difficulty Level : Medium
• Last Updated : 31 May, 2022

When adding two binary numbers by hand we keep the carry bits in mind and add it at the same time. But to do same thing in program we need a lot of checks. Recursive solution can be imagined as addition of carry and a^b (two inputs) until carry becomes 0.

Examples :

```Input : int x = 45, y = 45
Output : 90

Input : int x = 4, y = 78
Output : 82```

Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits.

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.

One important observation is, if (x & y) becomes 0, then result is x ^ y.

## C

 `// C program to do recursive addition``// of two integers``#include ` `int` `add(``int` `x, ``int` `y) {``    ``int` `keep = (x & y) << 1;``    ``int` `res = x^y;` `    ``// If bitwise & is 0, then there``    ``// is not going to be any carry.``    ``// Hence result of XOR is addition.``    ``if` `(keep == 0)``        ``return` `res;` `    ``add(keep, res);``}` `// Driver code``int` `main(){``    ``printf``(``"%d"``, add(15, 38));``    ``return` `0;``}`

## C++

 `// C++ program to do recursive addition``// of two integers``#include ``using` `namespace` `std;` `int` `add(``int` `x, ``int` `y) {``    ``int` `keep = (x & y) << 1;``    ``int` `res = x^y;` `    ``// If bitwise & is 0, then there``    ``// is not going to be any carry.``    ``// Hence result of XOR is addition.``    ``if` `(keep == 0)``        ``return` `res;` `    ``add(keep, res);``}` `// Driver code``int` `main(){``    ``cout<< add(15, 38);``    ``return` `0;``}` `// This code is contributed by jainlovely450.`

## Java

 `// Java program to do recursive addition``// of two integers``import` `java.io.*;` `class` `GFG {``    ` `    ``static` `int` `add(``int` `x, ``int` `y)``    ``{``        ``int` `keep = (x & y) << ``1``;``        ``int` `res = x^y;``    ` `        ``// If bitwise & is 0, then there``        ``// is not going to be any carry.``        ``// Hence result of XOR is addition.``        ``if` `(keep == ``0``)``            ``return` `res;``            ` `        ``return` `add(keep, res);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``System.out.println(add(``15``, ``38``));``    ``}``}` `// This code is contributed by Ajit.`

## Python3

 `    ` `# Python program to do recursive addition``# of two integers`` ` `def` `add(x, y):``    ``keep ``=` `(x & y) << ``1``;``    ``res ``=` `x^y;`` ` `    ``# If bitwise & is 0, then there``    ``# is not going to be any carry.``    ``# Hence result of XOR is addition.``    ``if` `(keep ``=``=` `0``):``        ``return` `res;`` ` `    ``return` `add(keep, res);` ` ` `# Driver code` `print``(add(``15``, ``38``));` `#  This code is contributed by Princi Singh`

## C#

 `// C# program to do recursive``// addition of two integers``using` `System;` `class` `GFG {``    ` `    ``static` `int` `add(``int` `x, ``int` `y)``    ``{``        ``int` `keep = (x & y) << 1;``        ``int` `res = x^y;``    ` `        ``// If bitwise & is 0, then there``        ``// is not going to be any carry.``        ``// Hence result of XOR is addition.``        ``if` `(keep == 0)``            ``return` `res;``            ` `        ``return` `add(keep, res);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``Console.Write(add(15, 38));``    ``}``}` `// This code is contributed by Smitha.`

## PHP

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## Javascript

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Output:

`53`

Time Complexity : O(logn)

Auxiliary Space: O(logn)

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