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.
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.
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