'How to determine if a number is odd in JavaScript
Can anyone point me to some code to determine if a number in JavaScript is even or odd?
Solution 1:[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 ); };
Solution 2:[2]
You could do something like this:
function isEven(value){
if (value%2 == 0)
return true;
else
return false;
}
Solution 3:[3]
function isEven(x) { return (x%2)==0; }
function isOdd(x) { return !isEven(x); }
Solution 4:[4]
Do I have to make an array really large that has a lot of even numbers
No. Use modulus (%). It gives you the remainder of the two numbers you are dividing.
Ex. 2 % 2 = 0 because 2/2 = 1 with 0 remainder.
Ex2. 3 % 2 = 1 because 3/2 = 1 with 1 remainder.
Ex3. -7 % 2 = -1 because -7/2 = -3 with -1 remainder.
This means if you mod any number x by 2, you get either 0 or 1 or -1. 0 would mean it's even. Anything else would mean it's odd.
Solution 5:[5]
This can be solved with a small snippet of code:
function isEven(value) {
return !(value % 2)
}
Hope this helps :)
Solution 6:[6]
In ES6:
const isOdd = num => num % 2 == 1;
Solution 7:[7]
With bitwise, codegolfing:
var isEven=n=>(n&1)?"odd":"even";
Solution 8:[8]
Like many languages, Javascript has a modulus operator %, that finds the remainder of division. If there is no remainder after division by 2, a number is even:
// this expression is true if "number" is even, false otherwise
(number % 2 == 0)
Similarly, if there is a remainder of 1 after division by 2, a number is odd:
// this expression is true if "number" is odd, false otherwise
(number % 2 == 1)
This is a very common idiom for testing for even integers.
Solution 9:[9]
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 .
Solution 10:[10]
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
Solution 11:[11]
if (X % 2 === 0){
} else {
}
Replace X with your number (can come from a variable). The If statement runs when the number is even, the Else when it is odd.
If you just want to know if any given number is odd:
if (X % 2 !== 0){
}
Again, replace X with a number or variable.
Solution 12:[12]
<script>
function even_odd(){
var num = document.getElementById('number').value;
if ( num % 2){
document.getElementById('result').innerHTML = "Entered Number is Odd";
}
else{
document.getElementById('result').innerHTML = "Entered Number is Even";
}
}
</script>
</head>
<body>
<center>
<div id="error"></div>
<center>
<h2> Find Given Number is Even or Odd </h2>
<p>Enter a value</p>
<input type="text" id="number" />
<button onclick="even_odd();">Check</button><br />
<div id="result"><b></b></div>
</center>
</center>
</body>
Solution 13:[13]
Subtract 2 to it recursively until you reach either -1 or 0 (only works for positive integers obviously) :)
Solution 14:[14]
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.
Solution 15:[15]
Just executed this one in Adobe Dreamweaver..it works perfectly. i used if (isNaN(mynmb))
to check if the given Value is a number or not, and i also used Math.abs(mynmb%2) to convert negative number to positive and calculate
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
</head>
<body bgcolor = "#FFFFCC">
<h3 align ="center"> ODD OR EVEN </h3><table cellspacing = "2" cellpadding = "5" bgcolor="palegreen">
<form name = formtwo>
<td align = "center">
<center><BR />Enter a number:
<input type=text id="enter" name=enter maxlength="10" />
<input type=button name = b3 value = "Click Here" onClick = compute() />
<b>is<b>
<input type=text id="outtxt" name=output size="5" value="" disabled /> </b></b></center><b><b>
<BR /><BR />
</b></b></td></form>
</table>
<script type='text/javascript'>
function compute()
{
var enter = document.getElementById("enter");
var outtxt = document.getElementById("outtxt");
var mynmb = enter.value;
if (isNaN(mynmb))
{
outtxt.value = "error !!!";
alert( 'please enter a valid number');
enter.focus();
return;
}
else
{
if ( mynmb%2 == 0 ) { outtxt.value = "Even"; }
if ( Math.abs(mynmb%2) == 1 ) { outtxt.value = "Odd"; }
}
}
</script>
</body>
</html>
Solution 16:[16]
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
Solution 17:[17]
Using % will help you to do this...
You can create couple of functions to do it for you... I prefer separte functions which are not attached to Number in Javascript like this which also checking if you passing number or not:
odd function:
var isOdd = function(num) {
return 'number'!==typeof num ? 'NaN' : !!(num % 2);
};
even function:
var isEven = function(num) {
return isOdd(num)==='NaN' ? isOdd(num) : !isOdd(num);
};
and call it like this:
isOdd(5); // true
isOdd(6); // false
isOdd(12); // false
isOdd(18); // false
isEven(18); // true
isEven('18'); // 'NaN'
isEven('17'); // 'NaN'
isOdd(null); // 'NaN'
isEven('100'); // true
Solution 18:[18]
A more functional approach in modern javascript:
const NUMBERS = "nul one two three four five six seven ocho nueve".split(" ")
const negate = f=> (...args)=> !f(...args)
const isOdd = n=> NUMBERS[n % 10].indexOf("e")!=-1
const isEven = negate(isOdd)
Solution 19:[19]
One liner in ES6 just because it's clean.
const isEven = (num) => num % 2 == 0;
Solution 20:[20]
Many people misunderstand the meaning of odd
isOdd("str")should be false.
Only an integer can be odd.isOdd(1.223)andisOdd(-1.223)should be false.
A float is not an integer.isOdd(0)should be false.
Zero is an even integer (https://en.wikipedia.org/wiki/Parity_of_zero).isOdd(-1)should be true.
It's an odd integer.
Solution
function isOdd(n) {
// Must be a number
if (isNaN(n)) {
return false;
}
// Number must not be a float
if ((n % 1) !== 0) {
return false;
}
// Integer must not be equal to zero
if (n === 0) {
return false;
}
// Integer must be odd
if ((n % 2) !== 0) {
return true;
}
return false;
}
JS Fiddle (if needed): https://jsfiddle.net/9dzdv593/8/
1-liner
Javascript 1-liner solution. For those who don't care about readability.
const isOdd = n => !(isNaN(n) && ((n % 1) !== 0) && (n === 0)) && ((n % 2) !== 0) ? true : false;
Solution 21:[21]
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;
}
Solution 22:[22]
How about this...
var num = 3 //instead get your value here
var aa = ["Even", "Odd"];
alert(aa[num % 2]);
Solution 23:[23]
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.
Solution 24:[24]
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;
}
Solution 25:[25]
By using ternary operator, you we can find the odd even numbers:
var num = 2;
result = (num % 2 == 0) ? 'even' : 'odd'
console.log(result);Solution 26:[26]
Another example using the filter() method:
let even = arr.filter(val => {
return val % 2 === 0;
});
// even = [2,4,6]
Solution 27:[27]
So many answers here but i just have to mention one point.
Normally it's best to use the modulo operator like % 2 but you can also use the bitwise operator like & 1. They both would yield the same outcome. However their precedences are different. Say if you need a piece of code like
i%2 === p ? n : -n
it's just fine but with the bitwise operator you have to do it like
(i&1) === p ? n : -n
So there is that.
Solution 28:[28]
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
Solution 29:[29]
A Method to know if the number is odd
let numbers = [11, 20, 2, 5, 17, 10];
let n = numbers.filter((ele) => ele % 2 != 0);
console.log(n);Sources
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