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

You can use a for statement and a conditional to determine if a number or series of numbers is odd:

for (var i=1; i<=5; i++) 
if (i%2 !== 0) {
    console.log(i)
}

This will print every odd number between 1 and 5.

Answer 2

One liner in ES6 just because it's clean.

const isEven = (num) => num % 2 == 0;

Answer 3

When you need to test if some variable is odd, you should first test if it is integer. Also, notice that when you calculate remainder on negative number, the result will be negative (-3 % 2 === -1).

function isOdd(value) {
  return typeof value === "number" && // value should be a number
    isFinite(value) &&                // value should be finite
    Math.floor(value) === value &&    // value should be integer
    value % 2 !== 0;                  // value should not be even
}

If Number.isInteger is available, you may also simplify this code to:

function isOdd(value) {
  return Number.isInteger(value)      // value should be integer
    value % 2 !== 0;                  // value should not be even
}

Note: here, we test value % 2 !== 0 instead of value % 2 === 1 is because of -3 % 2 === -1. If you don't want -1 pass this test, you may need to change this line.

Here are some test cases:

isOdd();         // false
isOdd("string"); // false
isOdd(Infinity); // false
isOdd(NaN);      // false
isOdd(0);        // false
isOdd(1.1);      // false
isOdd("1");      // false
isOdd(1);        // true
isOdd(-1);       // true