JavaScript style: don't use wrapper objects for primitive types

Question

In the Google JavaScript style guide, it says not to use wrapper objects for primitive types. It says it's "dangerous" to do so. To prove its point, it uses the example:

var x = new Boolean(false);
if (x) {
  alert('hi');  // Shows 'hi'.
}

OK, I give up. Why is the if code being executed here?

Because `Boolean(false) === !(new Boolean(false))`. And `Boolean(false) === new Boolean(false).valueOf()`. — katspaugh, Jul 29 '11 at 18:11

score 20 · Accepted Answer · answered Jul 29 '11 at 18:06

20

Because every variable that is typeof Object is truthy and wrappers are objects.

answered Jul 29 '11 at 18:06

marc

6,103
1
28
33

So basically you're saying that x is _not_ being assigned the value false as I thought it was. It's being assigned an "object", which evaluates to true. – Bill Jul 29 '11 at 18:16
Yes, an object that wraps the value, but as it is an object, it is truthy. – marc Jul 30 '11 at 16:21
1

`typeof null` -> "object", `!!null` -> false. Correct me if I'm wrong, but I believe nulls are falsy objects and a rather important exception to the above stated absolute rule. – Patrick Apr 03 '13 at 20:40
@Patrick `null` isn't an object. See [this](http://stackoverflow.com/a/18808270/894284). – Matt Fenwick Mar 05 '14 at 23:15
@MattFenwick I should have just said "nulls are falsy" and excluded the word "objects". Otherwise, I believe my comment still makes sense in context. Just because you have an `x` such that `typeof x` yields `"object"` is not a valid way to check that `x` is truthy. – Patrick Mar 07 '14 at 18:24

score 12 · Answer 2 · edited Aug 02 '16 at 16:41

12

if(x) will run if x is truthy.

x is truthy if it's not falsey.

x is falsey if x is null, undefined, 0, "", false

So since new Boolean(false) is an Object and an Object is truthy, the block runs

edited Aug 02 '16 at 16:41

Matthias

13,607
9
44
60

answered Jul 29 '11 at 18:06

Raynos

166,823
56
351
396

score 2 · Answer 3 · answered Apr 27 '14 at 19:26

2

In the if(x) case, it's actually evaluating the default Boolean of the object named and not its value of false.

So be careful using Boolean objects instead of Boolean values. =)

answered Apr 27 '14 at 19:26

Collin Graves

2,207
15
11

score 0 · Answer 4 · edited Sep 07 '16 at 18:29

0

The following code uses a Boolean object. The Boolean object is false, yet console.log("Found") still executes because an object is always considered true inside a conditional statement. It doesn’t matter that the object represents false; it’s an object, so it evaluates to true.

var found = new Boolean(false);
if (found) 
{    console.log("Found");
       // this executes
}

edited Sep 07 '16 at 18:29

Machavity

30,841
27
92
100

answered Dec 07 '15 at 04:03

AasaanHai

1

JavaScript style: don't use wrapper objects for primitive types

4 Answers4