Why does arr.map(parseInt) not work in JavaScript?

Question

I feel like I'm doing some noobish overlooking of things. Can anyone help me understand why the following doesn't work

["1.1", "2.2", "3.3"].map(parseInt);
//=> [1, NaN, NaN]

This works though ???

["1.1", "2.2", "3.3"].map(function(num) {
  return parseInt(num);
});
//=> [1, 2, 3]

This seems to work too

["1.1", "2.2", "3.3"].map(Number);
//=> [1.1, 2.2, 3.3]

Note `["1.1", "1.1", "1.1", "2.2", "3.3"].map(parseInt);` → `[1, NaN, 1, 2, 3]` — p.s.w.g, Jan 30 '14 at 23:21

score 6 · Accepted Answer · edited May 23 '17 at 12:11

6

Array.prototype.map passes more than one arguments to the callback. In particular the second argument it passes is the index of the element being processed.

parseInt accepts more than one argument. In particular the second argument it accepts is the base of the numeric system you are converting from.

When the second element is processed you are calling parseInt("2.2", 1), and since 2 is not a valid digit in base 1 the result is NaN.

If you declare a "wrapper" callback around parseInt then these additional arguments are lost and everything seems to work correctly, although you should always pass the second argument to parseInt.

edited May 23 '17 at 12:11

Community

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answered Jan 30 '14 at 23:18

Jon

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Ah, I knew it was noobish overlooking! Thanks for the insight. So is there any harm in me using `.map(Number)` then? – Mulan Jan 30 '14 at 23:22
@naomik: No, `Number` is fine. – Jon Jan 30 '14 at 23:22
@naomik `.map(parseFloat)` works too. – p.s.w.g Jan 30 '14 at 23:23
2

@naomik `["1.1", "2.2", "3.3"].map(Math.floor)` may do what you were originally intending with `parseInt`. – Matt Jan 30 '14 at 23:24
Thanks, Matt. The float is an okay solution too. If I needed to force an int, `Math.floor` looks like a good function to use :) – Mulan Jan 30 '14 at 23:25

Why does arr.map(parseInt) not work in JavaScript?

1 Answers1