Simplest way to check if two integers have same sign?

Question

Which is the simplest way to check if two integers have same sign? Is there any short bitwise trick to do this?

Answer 1

if (x * y) > 0...

assuming non-zero and such.

Answer 2

As a technical note, bit-twiddly solutions are going to be much more efficient than multiplication, even on modern architectures. It's only about 3 cycles that you're saving, but you know what they say about a "penny saved"...

Answer 3

For any size of int with two's complement arithmetic:

#define SIGNBIT (~((unsigned int)-1 >> 1))
if ((x & SIGNBIT) == (y & SIGNBIT))
    // signs are the same

Answer 4

Better way using std::signbit as follows:

std::signbit(firstNumber) == std::signbit(secondNumber);

It also support other basic types (double, float, char etc).

Answer 5

Just off the top of my head...

int mask = 1 << 31;
(a & mask) ^ (b & mask) < 0;

Answer 6

if (a*b < 0) sign is different, else sign is the same (or a or b is zero)