How to check if a number is a Integer

2019-06-03 23:31 Posted in Learn with JavaScript

JavaScript does not discriminate integer and float type of numbers, all the numbers are stored as 64-bit float values, just like double type in Java. On the other hand, in actual usage, like Array indexes, bit-wise operations, they are based on 32-bit.

Checking with mod

Any Integers can be fully divided by 1, using this principle to check whether a number is a integer.

function isInteger(obj) {
    return obj%1 === 0
}
isInteger(3) // true
isInteger(3.3) // false　

This method seems to be handy, but can fall short and have unexpected behaviors when handling Strings and special values.

isInteger('') // true
isInteger('3') // true
isInteger(true) // true
isInteger([]) // true

Because of that, we need to check that if this was a legit number or not first, such as using typeof

function isInteger(obj) {
    return typeof obj === 'number' && obj%1 === 0
}
isInteger('') // false
isInteger('3') // false
isInteger(true) // false
isInteger([]) // false

Using `Math.floor()`, `Math.ceil()`, `Math.round()` to check

Integers are always equal to themselves after rounding, using this trait to check whether a number is a integer

Example with Math.floor()

function isInteger(obj) {
    return Math.floor(obj) === obj
}
isInteger(3) // true
isInteger(3.3) // false
isInteger('') // false
isInteger('3') // false
isInteger(true) // false
isInteger([]) // false

Using this method, it’s safe to pass down parameters such as Strings and booleans and even arrays, with less coding as well.

Using `parseInt()`

function isInteger(obj) {
    return parseInt(obj, 10) === obj
}
isInteger(3) // true
isInteger(3.3) // false
isInteger('') // false
isInteger('3') // false
isInteger(true) // false
isInteger([]) // false

Also works, but this method has a slyly fall back:

isInteger(1000000000000000000000) // false

Such weird behavior was because parseInt() will always convert the first parameter into a String before actually parsing integers, so this is not a good option when dealing with big numbers.

Using bit-wise operations

function isInteger(obj) {
    return (obj | 0) === obj
}
isInteger(3) // true
isInteger(3.3) // false
isInteger('') // false
isInteger('3') // false
isInteger(true) // false
isInteger([]) // false

Seems neat, and has great time efficiency too, but as mentioned before, it can only handle numbers within the range of 32-bit, so that the following code will return false.

isInteger(Math.pow(2, 32)) // false
``

Then again, we do not need to handle with such big numbers from day to day usage.

## Using `Number.isInteger()`

This is a `ES6` only work around, supported by the newest version of `Chrome` and `Firefox`

```js
Number.isInteger(3) // true
Number.isInteger(3.1) // false
Number.isInteger('') // false
Number.isInteger('3') // false
Number.isInteger(true) // false
Number.isInteger([]) // false

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