Are there special equality conditions relating to 0 in javascript?

Question

I have two samples of code that I don't understand why they evaluate the way they do

Sample one (these evaluate to true):

    {} + [] == 0
    {} + 1 == 1
    0/1 == 0

But then this will evaluate to NaN instead of 0:

    ({} + []) / ({} + 1)

Sample 2 (you can chain the first):

    1 == "1" == 1/1

But this will evaluate false even though choosing any two will evaluate true.

    0 == "0" == 0/1

try the first set with '===' instead of '==' then go read up on the difference. — rlemon, Apr 07 '14 at 22:14
`{} + []` is interpreted as `+[]`. The `{}` is a "block", not an object. `[]` is converted first to a string (`''`), then to a number (`0`). It's not doing addition, it's the unary `+` operator. Same with `{} + 1`. When the `{}` is parenthesis, like in `({} + [])`, then it's treated as an object, and it becomes string concatenation (`'[object Object]' + ''`). — gen_Eric, Apr 07 '14 at 22:15

score 0 · Answer 1 · answered Apr 07 '14 at 22:19

You got explanations for first sample, so here is about second:

1 == "1" == 1/1 -> true

Explained: 1 == "1" -> true, then true == 1/1 evaluates to 1 == 1 which is true

0 == "0" == 0/1

Explained: 0 == "0" -> true, then true == 0/1 evaluates to 1 == 0 which is false

That is why my example 0 == 1 == 0 in the comments, evaluates to true.

0 == 1 -> false, false == 0 -> true

score 0 · Answer 2 · edited May 23 '17 at 10:33

Sample one (these evaluate to true):
{} + [] == 0
{} + 1 == 1

Please understand the difference between your expressions and the following:

 ({} + []) == 0
 ({} + 1) == 1

Considering that the parenthesised expressions evaluate to "[object Object]" (and "[object Object]1" respectively), it will become clear why the division yields NaN.

Sample two is a different thing.

Are there special equality conditions relating to 0 in javascript?

Yes. 0 is the only falsy number, which makes a difference when comparing with booleans (where false will be cast to 0).

1 == "1" == 1/1

Check the evaluation of this:

(1 == "1") == 1/1 // casting the string to a number
(1 == 1) == 1
true == 1 // casting the boolean to a number
1 == 1
true

Since 1/1 is 1 and equality is commutative, the same holds for 1 == ("1" == 1/1)

But this will evaluate false even though choosing any two will evaluate true.
0 == "0" == 0/1

No "but":

(0 == "0") == 0/1 // again, casting the string to a number
(0 == 0) == 0
true == 0 // again, casting the boolean to a number
1 == 0
false

Again it doesn't matter whether we do (0=="0)==0 or 0==("0"==0)

Even without the typecasting, in plain boolean logic is clear that

(true == true) == true

will have a different result than

(false == false) == false

JavaScript does not support chained comparison operators, the operations will be implicitly nested. If you need to test three values for equality, do (false == false) && (false == false).

score -1 · Answer 3 · answered Apr 07 '14 at 22:14

-1

When you use the == operator you allow coercion. That makes values either "truthy" or "falsy" depending on what operands are involved. Try it with the === operator to see what happens.

answered Apr 07 '14 at 22:14

user1329482

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3
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as I hit enter on my comment I saw your answer crop up :P nice. – rlemon Apr 07 '14 at 22:14
1

Using `===` for sample 1 won't make a difference actually – Bergi Apr 07 '14 at 22:21
As might be true, but I'm trying to figure out why someone would compare either an empty object or empty array to 1. Even in a loosely-typed language a programmer has to show some discipline--in fact, I'd say particularly in a loosely-typed language. – user1329482 Apr 08 '14 at 10:24

Are there special equality conditions relating to 0 in javascript?

3 Answers3