Question

A friend and I are going back and forth with brain-teasers and I have no idea how to solve this one. My assumption is that it’s possible with some bitwise operators, but not sure.

Answer 1

In C, with bitwise operators:

#include<stdio.h>  int add(int x, int y) {     int a, b;     do {         a = x & y;         b = x ^ y;         x = a << 1;         y = b;     } while (a);     return b; }   int main( void ){     printf( '2 + 3 = %d', add(2,3));     return 0; } 

XOR (x ^ y) is addition without carry. (x & y) is the carry-out from each bit. (x & y) << 1 is the carry-in to each bit.

The loop keeps adding the carries until the carry is zero for all bits.