isInt() JavaScript function

22 May 2006   23 comments   Web development

Mind That Age!

This blog post is 12 years old! Most likely, its content is outdated. Especially if it's technical.

Here's my take on a function that determines if a variable is an integer or not before or after it's been recast to an int. The functionality is very similar to Python's 'isdigit()' function which works like this:

>>> assert "1".isdigit()
>>> assert not "1.0".isdigit()

However, my Javascript function should return true when the input is an integer or a string of an integer. Here it goes:

function isInt(x) {
   var y = parseInt(x, 10);
   return !isNaN(y) && x == y && x.toString() == y.toString();
}
assert(isInt("1"));
assert(isInt(1));
assert(!isInt("1a"));
assert(!isInt("1.0"));

You can see it in action here.
To be honest, I'm writing about this here just to not forget it the next time I need a similar function. Sorry to cause any interweb-noise.

UPDATE

Corrected whitespacing and made a jsFiddle link.

Comments

Manuel SIMONOT

Thanx good idea i think. But are you sure the second part of the condition is useful?

Not sure:p

Peter Bengtsson

The second part is necessary so that "1.0" yields a false.

Anonymous

Indeed. My mistake:s

Anonymous

d

Kal

Am I allowed to use this function as long as I give you credit for it in my source code?

Okonomiyaki3000

how about this:

String.prototype.isInt = function()
{
var re = new RegExp("^[\d]+$");
return this.match(re);
}

marchaos

that will fail for negative numbers.

Anonymous

assert -> alert

ankit

Thanks the original function worked for me.

scott christopher

I tend to use:

function isInt (i) {
return (i % 1) == 0;
}

Peter Bengtsson

This fails on floating point numbers. isInt("1.1") is false but isInt("1.0") is true which it's not supposed to.

Stuart Thiel

Thanks, used and credited.

onli

Thank you.

halbesbit

sorry i post the false code :/, delete it please.
isInt=function (i) { return ((i % 1) == 0)? i:false; }

Peter Bengtsson

Thanks. That's what scott suggested above. His approach returns only true or false.

msangel

function isInteger(s){
return (s%(parseInt(s)/Number(s)))===0;
}

is perfect solution (=

gonzoyumo

except when you want to consider "0" is an integer (which does)

Tex

Manuel SIMONOT feeling was right, there is useless parts. But he picked the really wrong one...

function isInt(n) {
return n.toString()==parseInt(n).toString();
}
// 'a' != 'NaN' so 'a' is not an Int
// '1.0' != '1' so '1.0' is not an Int

Peter want "1.0" to be false. So n%1==0 is the wrong test for that. But this request is, in a way, absurd. "1.0" is findable as being a Float. 1.0 can't. It is parsed from Float to Int before being send to the function. Displaying a 1.0 can only be done from a string. alert(1.0); give 1. And the test wrote in the article doesn't test Integer or Float, only String.

Aadit M Shah

The simplest form of isInt is as follows (for a number n):

parseInt(n) === n; //isInt

That's it - you don't need a function call! Similarly, we have the following:

parseInt(n) !== n; //isNotInt
parseFloat(n) === n; //isFloat
parseFloat(n) !== n; //isNotFloat

Anonymous

And how would you check this "03" ?

Bobinours

It depend of the version of EcmaScript and the browser.

"If the input string begins with "0", radix is eight (octal). This feature is non-standard, and some implementations deliberately do not support it (instead using the radix 10)."

"The ECMAScript 5 specification of the function parseInt no longer allows implementations to treat Strings beginning with a 0 character as octal values."

(Source : https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/parseInt )

So... For this reason, you should always specify a radix when using parseInt().

parseInt("03", 10);

Anonymous

Why this:

var n = 3.0;
alert(parseInt(n)===n);

alerts "true"?

Peter Bengtsson

Python does the same:

>>> 3.0 == 3
True

But in python you can do:

>>> 3 is 3
True
>>> 3.0 is 3.0
True
>>> 3.0 is 3
False

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