Parse an integer (and only an integer) in JavaScript

Question

Prompt as it is possible to translate a string to number so that only any variant except for the integer produced an error.

func('17') = 17;
func('17.25') = NaN
func(' 17') = NaN
func('17test') = NaN
func('') = NaN
func('1e2') = NaN
func('0x12') = NaN

ParseInt does not work, because it does not work correctly.

ParseInt('17') = 17;
ParseInt('17.25') = 17 // incorrect
ParseInt(' 17') = NaN
ParseInt('17test') = 17 // incorrect
ParseInt('') = NaN
ParseInt('1e2') = 1 // incorrect

And most importantly: for the function to work in IE, Chrome and other browsers!!!

Number() will get you close, but do you want to discard decimal places or do you really want to reject them as NaN? — Christopher
– Christopher, Commented Apr 14, 2017 at 19:27

gyre · Accepted Answer · 2017-04-14 19:38:05Z

14

You can use a regular expression and the ternary operator to reject all strings containing non-digits:

function intOrNaN (x) {
  return /^\d+$/.test(x) ? +x : NaN
}

console.log([
  '17', //=> 17
  '17.25', //=> NaN
  ' 17', //=> NaN
  '17test', //=> NaN
  '', //=> NaN
  '1e2', //=> NaN
  '0x12' //=> NaN
].map(intOrNaN))

edited Apr 14, 2017 at 19:38

answered Apr 14, 2017 at 19:31

gyre

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2 Comments

Wessam El Mahdy Over a year ago

Why did you add + before x?

gyre Over a year ago

@WessamElMahdy The unary + operator casts the value of x to a number (hence strings of digits are reinterpreted as a base-10 numeric literal)

Linus Thiel · Accepted Answer · 2017-04-14 19:31:26Z

3

This would satisfy all your tests:

function func (str) {
  var int = parseInt(str, 10);
  return str == str.trim() && str == int ? int : NaN
}

answered Apr 14, 2017 at 19:31

Linus Thiel

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