Check If Two Integers Have The Same Sign

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 &

Solution 1:

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

bool sameSign = (n1 * n2) > 0

Solution 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 <

Solution 3:

Use bitwise xor

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

Solution 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)

Solution 5:

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