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.
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.
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);
}
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()
.
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
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
.
I received a few more interesting solutions via Twitter, check them out.
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()
.