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

Fewer characters of code, but might underflow for very small numbers:

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

Note: As @tsh correctly mentioned, an overflow with an intermediate result of Infinity or -Infinity does work. But an underflow with an intermediate result of +0 or -0 will fail, because +0 is not bigger than 0.

or without underflow, but slightly larger:

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