How to determine if a number is odd in JavaScript

Question

Can anyone point me to some code to determine if a number in JavaScript is even or odd?

Answer 1

With bitwise, codegolfing:

var isEven=n=>(n&1)?"odd":"even";

Answer 2

Using % will help you to do this...

You can create couple of functions to do it for you... I prefer separte functions which are not attached to Number in Javascript like this which also checking if you passing number or not:

odd function:

var isOdd = function(num) {
  return 'number'!==typeof num ? 'NaN' : !!(num % 2);
};

even function:

var isEven = function(num) {
  return isOdd(num)==='NaN' ? isOdd(num) : !isOdd(num);
};

and call it like this:

isOdd(5); // true
isOdd(6); // false
isOdd(12); // false
isOdd(18); // false
isEven(18); // true
isEven('18'); // 'NaN'
isEven('17'); // 'NaN'
isOdd(null); // 'NaN'
isEven('100'); // true

Answer 3

in ES6:

const isOdd = num => num % 2 == 1;

Answer 4

A more functional approach in modern javascript:

const NUMBERS = "nul one two three four five six seven ocho nueve".split(" ")

const negate = f=> (...args)=> !f(...args)
const isOdd  = n=> NUMBERS[n % 10].indexOf("e")!=-1
const isEven = negate(isOdd)

Answer 5

This can be solved with a small snippet of code:

function isEven(value) {
    return !(value % 2)
}

Hope this helps :)

Answer 6

if (X % 2 === 0){
} else {
}

Replace X with your number (can come from a variable). The If statement runs when the number is even, the Else when it is odd.

If you just want to know if any given number is odd:

if (X % 2 !== 0){
}

Again, replace X with a number or variable.