Check if two integers have the same sign

Question

I'm searching for an efficient way to check if two numbers have the same sign.

Basically I'm searching for a more elegant way than this:

var n1 = 1;
var n2 = -1;

( (n1 > 0 && n2 > 0) || (n1<0 && n2 < 0) )? console.log("equal sign"):console.log("different sign");

A solution with bitwise operators would be fine too.

Answer 1

You can multiply them together; if they have the same sign, the result will be positive.

bool sameSign = (n1 * n2) > 0

Answer 2

Fewer characters of code, but might overflow:

n1*n2 > 0 ? console.log("equal sign") : console.log("different sign or zero");

or without integer overflow, but slightly larger:

(n1>0) == (n2>0) ? console.log("equal sign") : console.log("different sign");

if you consider 0 as positive the > should be replaced with <