Testing whether a value is odd or even
JavascriptNumbersJavascript Problem Overview
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.
Javascript Solutions
Solution 1 - Javascript
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.
Solution 2 - Javascript
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.
Solution 3 - Javascript
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*[02468]$/.test(n);
}
function isOdd(n) {
return /^-?\d*[13579]$/.test(n);
}
Solution 4 - Javascript
Note: there are also negative numbers.
function isOddInteger(n)
{
return isInteger(n) && (n % 2 !== 0);
}
where
function isInteger(n)
{
return n === parseInt(n, 10);
}
Solution 5 - Javascript
Why not just do this:
function oddOrEven(num){
if(num % 2 == 0)
return "even";
return "odd";
}
oddOrEven(num);
Solution 6 - Javascript
To complete Robert Brisita's bit test .
if ( ~i & 1 ) {
// Even
}
Solution 7 - Javascript
var isOdd = x => Boolean(x % 2);
var isEven = x => !isOdd(x);
Solution 8 - Javascript
var isEven = function(number) {
// Your code goes here!
if (number % 2 == 0){
return(true);
}
else{
return(false);
}
};
Solution 9 - Javascript
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);
alert(isEven(8)); //true
alert(isEven(9)); //false
alert(isEven(-8)); //true
Solution 10 - Javascript
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!
Solution 11 - Javascript
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
Solution 12 - Javascript
Different way:
var isEven = function(number) {
// Your code goes here!
if (((number/2) - Math.floor(number/2)) === 0) {return true;} else {return false;};
};
isEven(69)
Solution 13 - Javascript
Otherway using strings because why not
function isEven(__num){
return String(__num/2).indexOf('.') === -1;
}
Solution 14 - Javascript
if (testNum == 0);
else if (testNum % 2 == 0);
else if ((testNum % 2) != 0 );
Solution 15 - Javascript
Maybe this? if(ourNumber % 2 !== 0)
Solution 16 - Javascript
var num = someNumber
isEven;
parseInt(num/2) === num/2 ? isEven = true : isEven = false;
Solution 17 - Javascript
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);
}
}
Solution 18 - Javascript
Check if number is even in a line of code:
var iseven=(_)=>_%2==0
Solution 19 - Javascript
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');
}
}
Solution 20 - Javascript
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)
Solution 21 - Javascript
This one is more simple!
var num = 3 //instead get your value here
var aa = ["Even", "Odd"];
alert(aa[num % 2]);
Solution 22 - Javascript
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
Solution 23 - Javascript
if (i % 2) {
return odd numbers
}
if (i % 2 - 1) {
return even numbers
}