# Checking whether a value is an integer in JavaScript

[2014-05-26] dev, javascript, jslang

Integers lead an odd life in JavaScript. In the ECMAScript specification, they only exist conceptually: All numbers are always floating point and integers are ranges of numbers without decimal fractions (for details, consult “Integers in JavaScript” in “Speaking JavaScript”). In this blog post, I explain how to check whether a value is an integer.

## ECMAScript 5  #

There are many ways in which you could implement this check. At this moment, you may want to take a break and try to write your own solution: a function isInteger(x) that returns true if x is an integer and false, otherwise.

Let’s look at a few examples.

### Checking via the remainder operator  #

One can use the remainder operator (%) to express the fact that a number is an integer if the remainder of dividing it by 1 is 0.

function isInteger(x) {
return x % 1 === 0;
}

I like this solution, because it is quite self-descriptive. It usually works as expected:

> isInteger(17)
true
> isInteger(17.13)
false

You have to be careful with the remainder operator, because the first operand determines the sign of the result: if it is positive, the result is positive, if it is negative, the result is negative.

> 3.5 % 1
0.5
> -3.5 % 1
-0.5

However, we are checking for zero, so that’s not an issue here. One problem remains: this function can return true for non-numbers, because % coerces its operands to numbers:

> isInteger('')
true
> isInteger('33')
true
> isInteger(false)
true
> isInteger(true)
true

That can be easily fixed by adding a type check:

function isInteger(x) {
return (typeof x === 'number') && (x % 1 === 0);
}

### Checking via Math.round()#

A number is an integer if it remains the same after being rounded to the “closest” integer. Implemented as a check in JavaScript, via Math.round():

function isInteger(x) {
return Math.round(x) === x;
}

This function works as it should:

> isInteger(17)
true
> isInteger(17.13)
false

It also handles non-numbers correctly, because Math.round() always returns numbers and === only returns true if both operands have the same type.

> isInteger('')
false

If you wanted to make the code more explicit, you could add a type check (like we did in the previous solution). Furthermore, Math.floor() and Math.ceil() work just as well as Math.round().

### Checking via bitwise operators  #

Bitwise operators provide another way of converting a number to a “close” integer:

function isInteger(x) {
return (x | 0) === x;
}

This solution (along with other solutions based on bitwise operators) has one disadvantage: it can’t handle numbers beyond 32 bits.

> isInteger(Math.pow(2, 32))
false

### Checking via parseInt()#

parseInt() also converts numbers to integers and can be used similarly to Math.round(). Let’s find out whether that is a good idea.

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

Like the Math.round() solution, this implementation handles non-numbers well, but it does not correctly identify all numbers as integers:

> isInteger(1000000000000000000000)
false

Why? parseInt() coerces its first parameter to string before parsing digits. It is not a good choice for converting numbers to integers.

> parseInt(1000000000000000000000, 10)
1
> String(1000000000000000000000)
'1e+21'

Above, parseInt() stops parsing '1e+21' before the first non-digit, e, which is why it returns 1.

TODO

> parseInt(0.0000007, 10)
7
> String(0.0000007)
'7e-7'

### Other solutions  #

I received a few more interesting solutions via Twitter, check them out.

## ECMAScript 6  #

Complementing Math.round() et al., ECMAScript 6 provides an additional way of converting numbers to integers: Math.trunc(). That function removes a number’s decimal fraction:

> Math.trunc(4.1)
4
> Math.trunc(4.9)
4
> Math.trunc(-4.1)
-4
> Math.trunc(-4.9)
-4

Furthermore, ECMAScript 6 makes the task of checking for integers trivial, because it comes with a built-in function Number.isInteger().