I have that expression
if (a === Infinity && b === 0 || a === -Infinity && b === 0 || a === 0 && b === Infinity || a === 0 && b === -Infinity) {
return NaN
}
I want short it, but I have no idea how to do this
UPDATE
If it possible, I cant use isFinite(), how to shorten else?
You can use !isFinite() to test if it's either Infinity or -Infinity.
if ((!isFinite(a) && b === 0) || (!isFinite(b) && a === 0)) {
return NaN;
}
If a and b are number typed, then:
if (!(a*b || isFinite(a+b))) return NaN;
If your linter warns about use of global functions, then:
if (!(a*b || Number.isFinite(a+b))) return NaN;
If you can't use multiplication:
if (!(a && b || Number.isFinite(a+b))) return NaN;
Math.absing both and then checking that the minimum is 0 and the maximum is infinity should do it.
const nums = [a, b].map(Math.abs);
if (Math.min(...nums) === 0 && Math.max(...nums) === Infinity) {
return NaN;
}
Can also sort the array.
const nums = [a, b]
.map(Math.abs)
.sort(a, b) => a - b); // .sort() works too, but could be more confusing
if (nums[0] === 0 && nums[1] === Infinity)) {
return NaN;
}