# Testing whether a value is odd or even

I decided to create simple isEven and isOdd function with a very simple algorithm:

``````function isEven(n) {
n = Number(n);
return n === 0 || !!(n && !(n%2));
}

function isOdd(n) {
return isEven(Number(n) + 1);
}
``````

That is OK if n is with certain parameters, but fails for many scenarios. So I set out to create robust functions that deliver correct results for as many scenarios as I could, so that only integers within the limits of javascript numbers are tested, everything else returns false (including + and - infinity). Note that zero is even.

``````// Returns true if:
//
//    n is an integer that is evenly divisible by 2
//
// Zero (+/-0) is even
// Returns false if n is not an integer, not even or NaN
// Guard against empty string

(function (global) {

function basicTests(n) {

// Deal with empty string
if (n === '')
return false;

// Convert n to Number (may set to NaN)
n = Number(n);

// Deal with NaN
if (isNaN(n))
return false;

// Deal with infinity -
if (n === Number.NEGATIVE_INFINITY || n === Number.POSITIVE_INFINITY)
return false;

// Return n as a number
return n;
}

function isEven(n) {

// Do basic tests
if (basicTests(n) === false)
return false;

// Convert to Number and proceed
n = Number(n);

// Return true/false
return n === 0 || !!(n && !(n%2));
}
global.isEven = isEven;

// Returns true if n is an integer and (n+1) is even
// Returns false if n is not an integer or (n+1) is not even
// Empty string evaluates to zero so returns false (zero is even)
function isOdd(n) {

// Do basic tests
if (basicTests(n) === false)
return false;

// Return true/false
return n === 0 || !!(n && (n%2));
}
global.isOdd = isOdd;

}(this));
``````

Can anyone see any issues with the above? Is there a better (i.e. more accurate, faster or more concise without being obfuscated) version?

There are various posts relating to other languages, but I can't seem to find a definitive version for ECMAScript. Use modulus:

``````function isEven(n) {
return n % 2 == 0;
}

function isOdd(n) {
return Math.abs(n % 2) == 1;
}
``````

You can check that any value in Javascript can be coerced to a number with:

``````Number.isFinite(parseFloat(n))
``````

This check should preferably be done outside the `isEven` and `isOdd` functions, so you don't have to duplicate error handling in both functions. I prefer using a bit test:

``````if(i & 1)
{
// ODD
}
else
{
// EVEN
}
``````

This tests whether the first bit is on which signifies an odd number. Note: there are also negative numbers.

``````function isOddInteger(n)
{
return isInteger(n) && (n % 2 !== 0);
}
``````

where

``````function isInteger(n)
{
return n === parseInt(n, 10);
}
`````` How about the following? I only tested this in IE, but it was quite happy to handle strings representing numbers of any length, actual numbers that were integers or floats, and both functions returned false when passed a boolean, undefined, null, an array or an object. (Up to you whether you want to ignore leading or trailing blanks when a string is passed in - I've assumed they are not ignored and cause both functions to return false.)

``````function isEven(n) {
return /^-?\d*\$/.test(n);
}

function isOdd(n) {
return /^-?\d*\$/.test(n);
}
`````` To complete Robert Brisita's bit test .

``````if ( ~i & 1 ) {
// Even
}
`````` ``````var isEven = function(number) {
if (number % 2 == 0){
return(true);
}
else{
return(false);
}
};
`````` A simple modification/improvement of Steve Mayne answer!

``````function isEvenOrOdd(n){
if(n === parseFloat(n)){
return isNumber(n) && (n % 2 == 0);
}
return false;
}
``````

Note: Returns false if invalid! We just need one line of code for this!

Here a newer and alternative way to do this, using the new ES6 syntax for JS functions, and the one-line syntax for the `if-else` statement call:

``````const isEven = num => ((num % 2) == 0) ? true : false;

`````` Different way:

``````var isEven = function(number) {
if (((number/2) - Math.floor(number/2)) === 0) {return true;} else {return false;};
};

isEven(69)
`````` Otherway using strings because why not

``````function isEven(__num){
return String(__num/2).indexOf('.') === -1;
}
`````` ``````if (testNum == 0);
else if (testNum % 2  == 0);
else if ((testNum % 2) != 0 );
`````` Maybe this? if(ourNumber % 2 !== 0) To test whether or not you have a odd or even number, this also works.

``````const comapare = x => integer(checkNumber(x));

function checkNumber (x) {
if (x % 2 == 0) {
return true;
}
else if (x % 2 != 0) {
return false;
}
}

function integer (x) {
if (x) {
console.log('even');
}
else {
console.log('odd');
}
}
`````` Using modern javascript style:

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

const isOdd  = n=> NUMBERS[n % 10].indexOf("e")!=-1
const isEven = n=> isOdd(+n+1)
`````` Simple way to test whether a number is even using JavaScript

``````function isEven(x) {
//return true if even
if (x % 2 === 0) {
return true;
}
//return false otherwise
else {
return false
}
}
// or much simpler
function isEven(x) {
return x % 2 === 0;
}
`````` This one is more simple!

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

`````` ``````if (i % 2) {
return odd numbers
}

if (i % 2 - 1) {
return even numbers
}
`````` Why not just do this:

``````    function oddOrEven(num){
if(num % 2 == 0)
return "even";
return "odd";
}
oddOrEven(num);
`````` ``````var num = someNumber
isEven;
parseInt(num/2) === num/2 ? isEven = true : isEven = false;
`````` A few

``````x % 2 == 0; // Check if even

!(x & 1); // bitmask the value with 1 then invert.

((x >> 1) << 1) == x; // divide value by 2 then multiply again and check against original value

~x&1; // flip the bits and bitmask
`````` ``````function isEven(n) {return parseInt(n)%2===0?true:parseInt(n)===0?true:false}
``````

when 0/even wanted but

``````isEven(0) //true
isEven(1) //false
isEven(2) //true
isEven(142856) //true
isEven(142856.142857)//true
isEven(142857.1457)//false
``````

? ``````for(var a=0; a<=20;a++){
if(a%2!==0){
console.log("Odd number "+a);
}
}

for(var b=0; b<=20;a++){
if(b%2===0){
console.log("Even number "+b);
}
}``````