How to determine if a number is odd in JavaScript

Use the below code:

function isOdd(num) { return num % 2;}
console.log("1 is " + isOdd(1));
console.log("2 is " + isOdd(2));
console.log("3 is " + isOdd(3));
console.log("4 is " + isOdd(4));

1 represents an odd number, while 0 represents an even number.


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 ); };

Tags:

Javascript

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