Hello Folks,

Write a function Add() that returns sum of two integers. The function should not use any of the arithmetic operators (+, ++, –, -, … etc).

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.

`// 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 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`

`int`

`main()`

`{`

` `

`cout << Add(15, 32);`

` `

`return`

`0;`

`}`

**Output :**

47

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.

Thankyou