Detecting if a given number is an integer

Question

Hello, what seems to be a simple question turns out as a difficult task to me. I need this:

if (number == integer) 

if (1.589 == integer) // false
if (2 == integer) // true

Any clues?

Answer 1

num % 1 === 0

This will convert num to type Number first, so any value which can be converted to an integer will pass the test (e.g. '42', true).

If you want to exclude these, additionally check for

typeof num === 'number'

You could also use parseInt() to do this, ie

parseInt(num) == num

for an untyped check and

parseInt(num) === num

for a typed check.

Note that the tests are not equivalent: Checking via parseInt() will first convert to String, so eg true won't pass the check.

Also note that the untyped check via parseInt() will handle hexadecimal strings correctly, but will fail for octals (ie numeric strings with leading zero) as these are recognized by parseInt() but not by Number(). If you need to handle decimal strings with leading zeros, you'll have to specify the radix argument.

Answer 2

There is a javascript function called isNaN(val) which returns true if val is not a number.

If you want to use val as a number, you need to cast using parseInt() or parseFloat()

EDIT: oops. Corrected the error as mentioned in the comment

Answer 3

How about this:

if((typeof(no)=='number') && (no.toString().indexOf('.')==-1))

Answer 4

you could either make use of javas parsing capabilities or as well try out the modulo operator %...

Answer 5

Would this not work:

if (parseInt(number, 10) == number)
{
  alert(number + " is an integer.");
}

Answer 6

Someone has already done it for you.

Answer 7

if (Math.floor(x) == x)

Answer 8

You could use the formal definition of integer:

Math.floor(x) === x