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.

ts. We calculate (x & y) << 1 and add it to x ^ y to get the required

result.

#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

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;

}

int main()

{

printf("%d", Add(15, 32));

return 0;

}

o/p:- 47

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.

ts. We calculate (x & y) << 1 and add it to x ^ y to get the required

result.

#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

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;

}

int main()

{

printf("%d", Add(15, 32));

return 0;

}

o/p:- 47