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

Assuming 32 bit ints:

bool same = ((x ^ y) >> 31) != 1;

Slightly more terse:

bool same = !((x ^ y) >> 31);

Answer 2

Thinking back to my university days, in most machine representations, isn't the left-most bit of a integer a 1 when the number is negative, and 0 when it's positive?

I imagine this is rather machine-dependent, though.

Answer 3

assuming 32 bit

if(((x^y) & 0x80000000) == 0)

... the answer if(x*y>0) is bad due to overflow

Answer 4

(a ^ b) >= 0

will evaluate to 1 if the sign is the same, 0 otherwise.

Answer 5

int same_sign = !( (x >> 31) ^ (y >> 31) );

if ( same_sign ) ... else ...

Answer 6

What's wrong with

return ((x<0) == (y<0));  

?