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

Use the bitwise AND operator.

function oddOrEven(x) {
  return ( x & 1 ) ? "odd" : "even";
}

function checkNumber(argNumber) {
  document.getElementById("result").innerHTML = "Number " + argNumber + " is " + oddOrEven(argNumber);
}
 
checkNumber(17);

<div id="result" style="font-size:150%;text-shadow: 1px 1px 2px #CE5937;" ></div>

If you don't want a string return value, but rather a boolean one, use this:

var isOdd = function(x) { return x & 1; };
var isEven  = function(x) { return !( x & 1 ); };

Answer 2

You could do something like this:

function isEven(value){
    if (value%2 == 0)
        return true;
    else
        return false;
}

Answer 3

This is what I did

//Array of numbers
var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10,32,23,643,67,5876,6345,34,3453];
//Array of even numbers
var evenNumbers = [];
//Array of odd numbers
var oddNumbers = [];

function classifyNumbers(arr){
  //go through the numbers one by one
  for(var i=0; i<=arr.length-1; i++){
     if (arr[i] % 2 == 0 ){
        //Push the number to the evenNumbers array
        evenNumbers.push(arr[i]);
     } else {
        //Push the number to the oddNumbers array
        oddNumbers.push(arr[i]);
     }
  }
}

classifyNumbers(numbers);

console.log('Even numbers: ' + evenNumbers);
console.log('Odd numbers: ' + oddNumbers);

For some reason I had to make sure the length of the array is less by one. When I don't do that, I get "undefined" in the last element of the oddNumbers array.

Answer 4

I'd implement this to return a boolean:

function isOdd (n) {
    return !!(n % 2);
    // or ((n % 2) !== 0).
}

It'll work on both unsigned and signed numbers. When the modulus return -1 or 1 it'll get translated to true.

Non-modulus solution:

var is_finite = isFinite;
var is_nan = isNaN;

function isOdd (discriminant) {
    if (is_nan(discriminant) && !is_finite(discriminant)) {
        return false;
    }

    // Unsigned numbers
    if (discriminant >= 0) {
        while (discriminant >= 1) discriminant -= 2;

    // Signed numbers
    } else {
        if (discriminant === -1) return true;
        while (discriminant <= -1) discriminant += 2;
    }

    return !!discriminant;
}

Answer 5

A simple function you can pass around. Uses the modulo operator %:

var is_even = function(x) {
    return !(x % 2); 
}

is_even(3)
false
is_even(6)
true

Answer 6

function isEven(x) { return (x%2)==0; }
function isOdd(x) { return !isEven(x); }