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

Every odd number when divided by two leaves remainder as 1 and every even number when divided by zero leaves a zero as remainder. Hence we can use this code

  function checker(number)  {
   return number%2==0?even:odd;
   }

Answer 2

Subtract 2 to it recursively until you reach either -1 or 0 (only works for positive integers obviously) :)

Answer 3

By using ternary operator, you we can find the odd even numbers:

var num = 2;
result = (num % 2 == 0) ? 'even' : 'odd'
console.log(result);

Answer 4

How about this...

    var num = 3 //instead get your value here
    var aa = ["Even", "Odd"];

    alert(aa[num % 2]);

Answer 5

Use my extensions :

Number.prototype.isEven=function(){
     return this % 2===0;
};

Number.prototype.isOdd=function(){
     return !this.isEven();
}

then

var a=5; 
 a.isEven();

==False

 a.isOdd();

==True

if you are not sure if it is a Number , test it by the following branching :

if(a.isOdd){
    a.isOdd();
}

UPDATE :

if you would not use variable :

(5).isOdd()

Performance :

It turns out that Procedural paradigm is better than OOP paradigm . By the way , i performed profiling in this FIDDLE . However , OOP way is still prettiest .

Answer 6

this works for arrays:

function evenOrOdd(numbers) {
  const evenNumbers = [];
  const oddNumbers = [];
  numbers.forEach(number => {
    if (number % 2 === 0) {
      evenNumbers.push(number);
    } else {
      oddNumbers.push(number);
    }
  });

  console.log("Even: " + evenNumbers + "\nOdd: " + oddNumbers);
}

evenOrOdd([1, 4, 9, 21, 41, 92]);

this should log out: 4,92 1,9,21,41

for just a number:

function evenOrOdd(number) {
  if (number % 2 === 0) {
    return "even";
  }

  return "odd";
}

console.log(evenOrOdd(4));

this should output even to the console