parseInt is not returning NaN when there is a letter

Question

parseInt('5aab4') //5
// I expect the output is NaN.

My logic fails, due to parseInt. I assume that parseInt always returns NaN, when the input contains a letter?

*"I assume that parseInt always returns NaN, when the input contains a letter"* - Don't assume, verify: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/parseInt You also may find some responses here useful: https://stackoverflow.com/questions/175739/built-in-way-in-javascript-to-check-if-a-string-is-a-valid-number — David, Apr 25 '19 at 20:12
Indeed, from the documentation: "If parseInt encounters a character that is not a numeral in the specified radix, it ignores it and all succeeding characters and returns the integer value parsed up to that point" — , Apr 25 '19 at 20:15
@ScottMarcus. I don't think it's a duplicate. My question is not how parseInt() works rather another option to check if a string only contains a number. — Henok Tesfaye, Apr 25 '19 at 20:23
You make no mention of another option in your question. You say *I assume that parseInt always returns NaN, when the input contains a letter?* which indicates you are asking about how `parseInt()` works. In fact, no where in the entire question do you even use the words *"check if a string only contains a number"*. — Scott Marcus, Apr 25 '19 at 20:29

Pointy · Answer 1 · 2019-04-25T20:25:07.257

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No, parseInt() allows "garbage" (non-number) characters after some numeric characters. The parseFloat() function has the same "feature".

If you want to get a number, you can use Number("123") or simply the + unary operator. However those accept any JavaScript number, including ones with fractional parts or numbers in scientific notation. But those methods will fail and give NaN if the string input is not a valid numeric constant.

You could do something like

if (+someString === Math.floor(+someString))

I guess.

edit — a comment notes that you'd also want to check the degenerate case of an empty or all-space string too. A simple regular expression (/^\d+$/) followed by a sanity check that it's not 200 digits long (amonth possibly other things) is another alternative.

edited Apr 25 '19 at 20:25

answered Apr 25 '19 at 20:11

Pointy

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I have never known why it works that way, but it always has. – Pointy Apr 25 '19 at 20:12
So, do you have another better way, to check if the string only contains a number? – Henok Tesfaye Apr 25 '19 at 20:13
"I have never known why it works that way, but it always has." [Because that's the specified behavior.](https://www.ecma-international.org/ecma-262/9.0/index.html#sec-parseint-string-radix) – Jordan Running Apr 25 '19 at 20:16
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@JordanRunning yes of course that's true, but it was designed and implemented *before* the language semantics were codified in a specification. – Pointy Apr 25 '19 at 20:17
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`+' ' === 0`. JavaScript is *wonderful*. – AuxTaco Apr 25 '19 at 20:22
@AuxTaco yea good point, I generally just use a regex for checking integers – Pointy Apr 25 '19 at 20:24

score -1 · Answer 2 · answered Apr 25 '19 at 20:14

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If you were wanting to check NaNanother way would be to test it with isNaN(). isNaN() will return a boolean giving you a verification for a given input.

parseInt('5aab4') //5

isNaN('5aab4') // true

answered Apr 25 '19 at 20:14

Bryan

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it won't limit the input to just integers however – Pointy Apr 25 '19 at 20:16
But keep in mind that `isNaN('6e23')` and `isNaN(' ')` are false. – AuxTaco Apr 25 '19 at 20:19

parseInt is not returning NaN when there is a letter

2 Answers2