Assigning `{} + []` to a variable

Question

>> {} + []
0

So EmptyObject + EmptyArray evaluates to 0. (Disclaimer: I don't fully understand why this is.) Okay, now I want to assign this 0 to a variable! Seems pretty easy:

>> var a = {} + [];
undefined
>> a
"[object Object]"

Well, maybe not so easy. Googles a bit... Aha! {} in this context is actually a code block, not an empty object. So lemme make it look more like a code block:

>> var a = 
   {

   }
   + [];
undefined
>> a
"[object Object]"

...dang. Note to self: JavaScript ignores whitespace.

Is there any way I can assign {} + [] (or similar) to a variable, but still have that {} + [] evaluate to 0?

@NiettheDarkAbsol Because it would be funny. :P Do I have to have a better reason? — The Guy with The Hat, Mar 31 '14 at 18:41
This reminds me of this gem: https://www.destroyallsoftware.com/talks/wat — Lukas S., Mar 31 '14 at 18:42
@SergeyK That's actually where I first heard about `{} + [] = 0` — The Guy with The Hat, Mar 31 '14 at 18:43
@KevinB That's what I figured, given what the linked answer said. — The Guy with The Hat, Mar 31 '14 at 18:43
This may help: http://www.2ality.com/2012/01/object-plus-object.html — Matt Burland, Mar 31 '14 at 18:53
possible duplicate of [Why an empty Array type-converts to zero? +\[\]](http://stackoverflow.com/questions/3306453/why-an-empty-array-type-converts-to-zero) — Huangism, Mar 31 '14 at 18:57

Chuck · Accepted Answer · 2014-03-31T23:02:19.137

No, there is no way to assign that exact code to a variable and have it work out to 0, because the parsing of that code that gives 0 does not constitute an expression†, and only expressions can be assigned. The reason you're getting that 0 is because the braces are being essentially ignored as an empty statement and then you're getting the result of the expression + [] (that is, the unary + operator applied to an empty array). So in order to get the same result out, you'd need to alter the expression so that it just worked out to +[]. For example, you could do var a = ({}, + []) and that would give you the right answer (because the comma causes + [] to be evaluated separately).

† This is also the reason it gives a different answer when you try to assign it. That forces the code to be parsed as an expression, so the {} is no longer an empty block and the + is no longer the unary +, but instead it's the binary plus with an empty object and an empty array as operands.

i bow before you o' great God of Javascript! – Randy L Mar 31 '14 at 20:28 — Randy L, Mar 31 '14 at 20:28

score 3 · Answer 2 · answered Mar 31 '14 at 18:41

3

Try:

var a = eval("{} + []");

console.log(a); // returns (integer) 0

Though eval method isn't recommended to use

answered Mar 31 '14 at 18:41

nanobash

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*How* isn't the most important here, I think, but *why* is. – DontVoteMeDown Mar 31 '14 at 18:42
maybe the length of {} + the length of [] which is 0 since both are empty – Huangism Mar 31 '14 at 18:45
@Huangism Nope, it is not calculated with length because if we'd have `[2]` for instance the result will be `2` – nanobash Mar 31 '14 at 18:47
@ToKeN yea but [2] + 2 results in 22 and not 4, so [2] must of been a string – Huangism Mar 31 '14 at 18:48
@Huangism No, `eval("{} + [2] + 2");` gives `4` – nanobash Mar 31 '14 at 18:50
@Huangism Yes, without `eval` it gather result as a `string` – nanobash Mar 31 '14 at 18:52
Even without `eval`, `{} + [2] + 2` gives `4`. That's because it's treated as `+[2] + 2`, and `+[2]` evaluates to `2`. However, `[2] + 2` is different, this gives `22` because in that case `[2]` is treated as `"2"`. – Tim Goodman Mar 31 '14 at 19:03

Assigning `{} + []` to a variable

2 Answers2