How to check if a JavaScript number is a real, valid number?

Question

MY code is:

function isNumber(n){
return typeof n == 'number' && !isNaN(n);
}

window.onload=function(){
var a=0,b=1,c=2.2,d=-3,e=-4.4,f=10/3;
var shouldBeTrue=[a,b,c,d,e,f];

var aa="0",bb="1",cc="2.2",dd="-3",ee="-4.4",ff="10/3";
var shouldBeFalse=[aa,bb,cc,dd,ee,ff];

var aaa,bbb=true,ccc=false,ddd=document.getElementsByTagName('html');
var alsoTheseBeFalse=[aaa,bbb,ccc,ddd,""," ",,null,NaN];

for(var i=0;i<shouldBeTrue.length;i++)
    if(isNumber(shouldBeTrue[i]) != true) alert("x");
for(i=0;i<shouldBeFalse.length;i++)
    if(isNumber(shouldBeFalse[i]) != false) alert("x");
for(i=0;i<alsoTheseBeFalse.length;i++)
    if(isNumber(alsoTheseBeFalse[i]) != false) alert("x");
}

What else should I check against to ensure my function is 101% perfect in all ways? (also, if you know a better function please tell me)

which function are you looking to make perfect? You could start by giving your variables more meaningful names... — frenchie
– frenchie, Commented Dec 15, 2011 at 20:15
There were numeric cases you didn't check, but I cant make your function break (ie: g=2E30, gg=0/0) — Brian
– Brian, Commented Dec 15, 2011 at 20:19
@Brian, well 0/0=NaN (and I did check for that), but I am curious about the '2E30' thing because it returns true. Do you know why it returns true? — user1022373
– user1022373, Commented Dec 15, 2011 at 20:24
Side note, inside your test cases, you're using != false. This is not needed, because your isNumber function always return a boolean (true or false). Within if ( ... ), an expressions is always treathed as a boolean. if (isNumber(a) != false) is equivalent to if (isNumber(a) == true) is equivalent to if (isNumber(a)). — Rob W
– Rob W, Commented Dec 15, 2011 at 20:47

Rob W · Accepted Answer · 2011-12-15 20:38:43Z

24

If you want to check whether a number is a real number, you should also check whether it's finite:

function isNumber(n){
    return typeof n == 'number' && !isNaN(n) && isFinite(n);
 }

Another method (explanation below):

function isNumber(n){
    return typeof n == 'number' && !isNaN(n - n);
}

Update: Two expressions to validate a real number

Since JavaScript numbers are representing real numbers, the substraction operand on the same number should produce the zero value (additive identity). Numbers out of range should (and will) be invalid, NaN.

1        - 1        = 0    // OK
Infinity - Infinity = NaN  // Expected
NaN      - NaN      = NaN  // Expected
NaN      - Infinity = NaN

edited Dec 15, 2011 at 20:38

answered Dec 15, 2011 at 20:15

Rob W

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12 Comments

user1022373 Over a year ago

Well, your code is larger and do 1 more operation. The only way I would use it is if you to tell me a way that my current function would fail and yours not

Rob W Over a year ago

@TuxedoKnightChess Input the following value: Infinity. It passes the first two tests, while it's not a useful real number: isNumber(Infinity) returns true for your current function.

user1022373 Over a year ago

what would be the result for var n=2E30;?

RightSaidFred Over a year ago

@TuxedoKnightChess: There's a real easy way to find out.

Rob W Over a year ago

@TuxedoKnightChess The methods slightly differ, but the results are exactely equal.

|

Community · Accepted Answer · 2017-05-23 11:46:59Z

JS numbers can be among the following values:

Finite numbers
+Infinity and -Infinity
NaN

Then there also non-number values which are coercible to numbers, e.g. number objects. You might want to consider them numerical.

If you only want to test finite numbers, simply use Number.isFinite:

Number.isFinite(value)

var isNumber = Number.isFinite;
assert('isNumber(1)', true);
assert('isNumber(1.1)', true);
assert('isNumber(+0)', true);
assert('isNumber(-0)', true);
assert('isNumber(-1.1)', true);
assert('isNumber(Math.PI)', true);
assert('isNumber(1e300)', true);
assert('isNumber(+Infinity)', false);
assert('isNumber(-Infinity)', false);
assert('isNumber(NaN)', false);
assert('isNumber(null)', false);
assert('isNumber(undefined)', false);
assert('isNumber(true)', false);
assert('isNumber(false)', false);
assert('isNumber("123")', false);
assert('isNumber("foo")', false);
assert('isNumber(new Number(1))', false);
assert('isNumber([])', false);
assert('isNumber({})', false);
assert('isNumber(function(){})', false);
function assert(code, expected) {
  var result = eval(code);
  console.log('Test ' + (result===expected ? 'pass' : 'FAIL') + ': ', code, ' -> ', result);
}

If you want to include infinities, check the type and exclude NaN:

typeof value === "number" && !Number.isNaN(value)

function isNumber(value) {
  return typeof value === "number" && !Number.isNaN(value);
}
assert('isNumber(1)', true);
assert('isNumber(1.1)', true);
assert('isNumber(+0)', true);
assert('isNumber(-0)', true);
assert('isNumber(-1.1)', true);
assert('isNumber(Math.PI)', true);
assert('isNumber(1e300)', true);
assert('isNumber(+Infinity)', true);
assert('isNumber(-Infinity)', true);
assert('isNumber(NaN)', false);
assert('isNumber(null)', false);
assert('isNumber(undefined)', false);
assert('isNumber(true)', false);
assert('isNumber(false)', false);
assert('isNumber("123")', false);
assert('isNumber("foo")', false);
assert('isNumber(new Number(1))', false);
assert('isNumber([])', false);
assert('isNumber({})', false);
assert('isNumber(function(){})', false);
function assert(code, expected) {
  var result = eval(code);
  console.log('Test ' + (result===expected ? 'pass' : 'FAIL') + ': ', code, ' -> ', result);
}

If you want to consider number objects as numbers, you can unwrap them using

value = Number.valueOf.call(value); // throws if value was not a number object

function isNumber(value) {
  try { value = Number.prototype.valueOf.call(value); } catch(err) { }
  return Number.isFinite(value);
}
assert('isNumber(1)', true);
assert('isNumber(1.1)', true);
assert('isNumber(+0)', true);
assert('isNumber(-0)', true);
assert('isNumber(-1.1)', true);
assert('isNumber(Math.PI)', true);
assert('isNumber(1e300)', true);
assert('isNumber(+Infinity)', false);
assert('isNumber(-Infinity)', false);
assert('isNumber(NaN)', false);
assert('isNumber(null)', false);
assert('isNumber(undefined)', false);
assert('isNumber(true)', false);
assert('isNumber(false)', false);
assert('isNumber("123")', false);
assert('isNumber("foo")', false);
assert('isNumber(new Number(1))', true);
assert('isNumber([])', false);
assert('isNumber({})', false);
assert('isNumber(function(){})', false);
function assert(code, expected) {
  var result = eval(code);
  console.log('Test ' + (result===expected ? 'pass' : 'FAIL') + ': ', code, ' -> ', result);
}

If you want to include arbitrary values coercible to numbers, you can use the unary + to coerce.

value = +value; // throws if value was not number-coercible

There is also the isNaN function (not to be confused with Number.isNaN), which will first coerce and then compare with NaN. But be aware whitespace strings and null are coerced to +0, not NaN. So you might be interested in Validate decimal numbers in JavaScript - IsNumeric()

Jukka K. Korpela · Accepted Answer · 2011-12-15 20:16:43Z

0

It depends on what you wish to regard as a number. Your code classifies Infinity and -Infinity as numbers. If you don’t want that, replace !isNaN(n) by isFinite(n).

And your code classifies '42' (a string literal) as not being a number, due to the type check; but I supposed that’s intentional.

answered Dec 15, 2011 at 20:16

Jukka K. Korpela

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Comments

Ameer Khalaf · Accepted Answer · 2018-07-30 10:23:44Z

0

I use combination of parse and check for numbers.. as outlined below

function isNumber(inputValue){ return ((parseFloat(inputValue) ==0 || parseFloat(inputValue)) && !isNaN(inputValue)); };

hope this helps

answered Jul 30, 2018 at 10:23

Ameer Khalaf

11 silver badge

Comments

Sunil Jhamnani · Accepted Answer · 2020-02-11 15:30:45Z

0

For react or other framework users, possible solution could be:

const { onChange, value } = this.props;
return (
  <input
    type="text" // Note: I have kept this as text
    value={typeof value !== 'number' || isNaN(value) ? '' : value}
    className={classnames('input')}
    onChange={(event) =>
      onChange &&
      onChange(parseFloat(event.target.value))
    }
  />)

This works on safari as well.

Thanks.

answered Feb 11, 2020 at 15:30

Sunil Jhamnani

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Comments

RightSaidFred · Accepted Answer · 2011-12-18 17:10:59Z

-1

If you consider a number in its object wrapper to be a number, then it will fail with:

isNumber( new Number(123) )

Since a downvoter is having some comprehension troubles that couldn't be alleviated by a simple test, new Number(123) will return 'object' from the typeof test, and as such will not pass.

edited Dec 18, 2011 at 17:10

answered Dec 15, 2011 at 20:18

RightSaidFred

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2 Comments

RightSaidFred Over a year ago

@down-voter: Give a reason. Please tell me how new Number(123) will pass the isNumber test in the question.

Rob W Over a year ago

I did not downvote, but your code only shows an example of something that breaks the code, without proposing a solution. The OP did not ask how to break the function, but how to improve the function.

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