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3314

answers:

7

I recently discovered that 2 == [2] in JavaScript. As it turns out, this quirk has a couple of interesting consequences:

var a = [0, 1, 2, 3];
a[[2]] === a[2]; // this is true

Similarly, the following works:

var a = { "abc" : 1 };
a[["abc"]] === a["abc"]; // this is also true

Even stranger still, this works as well:

[[[[[[[2]]]]]]] == 2; // this is true too! WTF?

These behaviors seem consistent across all browsers.

Any idea why this is a language feature?

EDIT: Here's another insane consequence of this "feature":

[0] == false // true
if ([0]) { /* executes */ } // [0] is both true and false!

This was found by jimbojw of trephine and http://jimbojw.com fame.

+7  A: 

A array of one item can be treated as the item itself.

This is due to duck typing. Since "2" == 2 == [2] and possibly more.

Ólafur Waage
2 === [2] does not work.
Chetan Sastry
Hmm it appears so, my bad.
Ólafur Waage
because they dont match in type. in the first example, the left side is evaluated first, and they wind up matching in type.
Shawn Simon
Also, I don't think duck-typing is the correct word here. It is more to do with the implicit type conversion performed by `==` operator before comparing.
Chetan Sastry
this has nothing to do with duck typing but rather with weak typing, ie implicit type conversion
Christoph
@Chetan: what he said ;)
Christoph
What Chetan and Christoph said.
Tim Down
+7  A: 

It is because of the implicit type conversion of == operator.

[2] is converted to Number is 2 when compared with a Number. Try the unary + operator on [2].

> +[2]
2
Chetan Sastry
Others are saying [2] is converted to a string. `+"2"` is also the number 2.
dlamblin
+6  A: 

For the == case, this is why Doug Crockford recommends always using ===. It doesn't do any implicit type conversion.

For the examples with ===, the implicit type conversion is done before the equality operator is called.

Dan Hook
he uses === in his example.
Ólafur Waage
i think youre right. i was just reading that book on the train this mornign : )
Shawn Simon
+6  A: 
var a = [0, 1, 2, 3];
a[[2]] === a[2]; // this is true

On the right side of the equation, we have the a[2], which returns a number type with value 2. On the left, we are first creating a new array with a single object of 2. Then we are calling a[(array is in here)]. I am not sure if this evaluates to a string or a number. 2, or "2". Lets take the string case first. I believe a["2"] would create a new variable and return null. null !== 2. So lets assume it is actually implicitly converting to a number. a[2] would return 2. 2 and 2 match in type (so === works) and value. I think it is implicitly converting the array to a number because a[value] expects a string or number. It looks like number takes higher precedence.

On a side note, I wonder who determines that precedence. Is because [2] has a number as it's first item, so it converts to a number? Or is it that when passing an array into a[array] it tries to turn the array into a number first, then string. Who knows?

var a = { "abc" : 1 };
a[["abc"]] === a["abc"];

In this example, you are creating an object called a with a member called abc. The right side of the equation is pretty simple; it is equivalent to a.abc. This returns 1. The left side first creates a literal array of ["abc"]. You then search for a variable on the a object by passing in the newly created array. Since this expects a string, it converts the array into a string. This now evaluates to a["abc"], which equals 1. 1 and 1 are the same type (which is why === works) and equal value.

[[[[[[[2]]]]]]] == 2;

This is just an implicit conversion. === wouldn't work in this situation because there is a type mismatch.

Shawn Simon
The answer to your question about precedence: `==` applies `ToPrimitive()` to the array, which in turn invokes its `toString()` method, so what you actually compare is the number `2` to the string `"2"`; comparison between a string and a number is done by converting the string
Christoph
+55  A: 

You can look up the comparison algorithm in the ECMA-spec (relevant sections of ECMA-262, 3rd edition for your problem: 11.9.3, 9.1, 8.6.2.6).

If you translate the involved abstract algorithms back to JS, what happens when evaluating 2 == [2] is basically this:

2 === Number([2].valueOf().toString())

where valueOf() for arrays returns the array itself and the string-representation of a one-element array is the string representation of the single element.

This also explains the third example as [[[[[[[2]]]]]]].toString() is still just the string 2.

As you can see, there's quite a lot of behind-the-scene magic involved, which is why I generally only use the strict equality operator ===.

The first and second example are easier to follow as property names are always strings, so

a[[2]]

is equivalent to

a[[2].toString()]

which is just

a["2"]

Keep in mind that even numeric keys are treated as property names (ie strings) before any array-magic happens.

Christoph
+2  A: 

To add a little detail to the other answers... when comparing an Array to a Number, Javascript will convert the Array with parseFloat(array). You can try it yourself in the console (eg Firebug or Web Inspector) to see what different Array values get converted to.

parseFloat([2]); // 2
parseFloat([2, 3]); // 2
parseFloat(['', 2]); // NaN

For Arrays, parseFloat performs the operation on the Array's first member, and discards the rest.

Edit: Per Christoph's details, it may be that it is using the longer form internally, but the results are consistently identical to parseFloat, so you can always use parseFloat(array) as shorthand to know for sure how it will be converted.

eyelidlessness
+3  A: 
[0] == false // true
if ([0]) { /* executes */ } // [0] is both true and false!

That's interesting, it's not that [0] is both true and false, actually

[0] == true // false

It is javascript's funny way of processing if() operator.

Alexander Abramov
actually, it's the funny way `==` works; if you use an actual explicit cast (ie `Boolean([0])` or `!![0]`), you'll find that `[0]` will evaluate to `true` in boolean contexts as it should: in JS, any object is considered `true`
Christoph