Fastest way to check if two integers are on the same side of 0

Question

I need to check if two integers are on the same side of zero many times. I don\'t care if it\'s positive or negative, just that it\'s the same side... and performance is very im

Answer 1

Alternate answers

Compare the sign bit

return ((n >> 31) ^ (n2 >> 31) ) == 0 ? /* same */ : /* different */;

Alternate way of comparing sign bit

return (((int1 & 0x80000000) ^ (int2 & 0x80000000))) == 0 ? /* same */ : /* different */;

and I just verified but Op's code is wrong when int1 == int2. The following will always print different if they are the same.

if (int1 == 0 || int2 == 0) {
    // handle zero
} else if ((int1 ^ int2) < 0) {
    // same side
} else {
    // different side
}