Compute the minimum (min) or maximum (max) of two integers without branching

int x;  // we want to find the minimum of x and y
int y;   
int r;  // the result goes here 

r = y ^ ((x ^ y) & -(x < y)); // min(x, y)

On some rare machines where branching is very expensive and no condition move instructions exist, the above expression might be faster than the obvious approach, r = (x < y) ? x : y, even though it involves two more instructions. (Typically, the obvious approach is best, though.) It works because if x < y, then -(x < y) will be all ones, so r = y ^ (x ^ y) & ~0 = y ^ x ^ y = x. Otherwise, if x >= y, then -(x < y) will be all zeros, so r = y ^ ((x ^ y) & 0) = y. On some machines, evaluating (x < y) as 0 or 1 requires a branch instruction, so there may be no advantage.To find the maximum, use:

r = x ^ ((x ^ y) & -(x < y)); // max(x, y)

http://graphics.stanford.edu/~seander/bithacks.html#IntegerMinOrMax