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

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");

Answer 2

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

bool sameSign = (n1 * n2) > 0

Answer 3

Use bitwise xor

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

Answer 4

That based on how you define "same sign" for special values:

• does NaN, NaN have the same sign?

If your answer is "No", the answer is:

Math.sign(a) === Math.sign(b)

If your answer is "Yes", the answer is:

Object.is(Math.sign(a) + 0, Math.sign(b) + 0)

Answer 5

n = n1*n2;
if(n>0){ same sign }
else { different sign }

Answer 6

Simple easy:

bool sameSign = ((n1 < 0) == (n2 < 0));