Testing whether a value is odd or even

Question

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.

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);
}

Note: there are also negative numbers.

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

where

function isInteger(n)
{
   return n === parseInt(n, 10);
}

Why not just do this:

    function oddOrEven(num){
        if(num % 2 == 0)
            return "even";
        return "odd";
    }
    oddOrEven(num);

To complete Robert Brisita's bit test .

if ( ~i & 1 ) {
    // Even
}

var isOdd = x => Boolean(x % 2);
var isEven = x => !isOdd(x);