why does "[] == 0" return true while "[]" is true and "0" is false?

Question

If i execute the following lines in thw browser console

!![] //--- returns true
!!0 //--- returns false

I get that [] and 0 have different boolean values.

I don't understand why

[] == 0 //--- returns true

returns true.

What am I missing?

So many duplicates.. tldr; there are different conversions that occur. `!` applies [ToBoolean] [*while `==` uses different conversions*](https://es5.github.io/#x11.9.3) ([ToNumber], [ToPrimitive]) depending on the type of both operands. — user2864740, Aug 06 '15 at 11:05
The `!` operator always converts to boolean, but `==` does other conversions to turn the data into values that are possible to compare. — Guffa, Aug 06 '15 at 11:08
`[]` and `0` are both falsey values so both are equal to each other — hitesh1124, Aug 06 '15 at 11:19
See http://stackoverflow.com/questions/7615214/in-javascript-why-is-0-equal-to-false-but-when-tested-by-if-it-is-not-fals?lq=1 — user2864740, Aug 06 '15 at 11:21
@hitesh1124 That is not correct. As a counter example, `false == null` is *not* true. — user2864740, Aug 06 '15 at 11:21
For more clarifications check this out! http://stackoverflow.com/questions/5491605/empty-arrays-seem-to-equal-true-and-false-at-the-same-time — Vibhesh Kaul, Aug 06 '15 at 11:28
just to do your head in a little more ... `[x] == x` but `[x] != [x]` :p — Jaromanda X, Aug 06 '15 at 11:40
@JaromandaX Not universally true. As a counter example, `[undefined] == undefined` is *not* true. — user2864740, Aug 06 '15 at 11:41
you may want to read this http://stackoverflow.com/q/7202157/995714 — phuclv, Aug 06 '15 at 12:10

score 6 · Accepted Answer · answered Aug 06 '15 at 11:32

Remember that array is object and 0 is number.

And as "user2864740" told..

1) When you doing

!![] //--- returns true
!!0 //--- returns false

You are performing so called "ToBoolean" convertion

https://es5.github.io/#x9.2

Number

The result is false if the argument is +0, −0, or NaN; otherwise the result is true.

Object ( our [] )

always true

2) But when you using == you performing so called "Equality Comparison"

https://es5.github.io/#x11.9.3

Here thins a little bit complicated but to understand what happens you have to remember that == do a type coercion ( so you can compare oranges to apples :) )

First of all compiler converts [] to some primitive type.

If Type(x) is either String or Number and Type(y) is Object, return the result of the comparison x == ToPrimitive(y).

How ToPrimitive works is a matter of an article :), but's it easy to remember that closet primitive type to array is string. Array will be converted to empty string.

[].toString() === ""

So now we need to compare empty string and number 0

"" == 0   // true

Hmmm. So it's true. But why is that? Remember that when you compare with "Equality Comparison" number and string

If Type(x) is Number and Type(y) is String, return the result of the comparison x == ToNumber(y).

So let's try to convert empty string to number

Number("") === 0

And in the end

0 === 0

I hope that's explains something :)

score 1 · Answer 2 · answered Aug 06 '15 at 11:10

1

JavaScript is probably converting the array to a number:
!!Number([]) // false
Number([]) == 0 // true

answered Aug 06 '15 at 11:10

G_hi3

588
5
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is that because the left side is converted to the type of the right side ?? – Marwan Aug 06 '15 at 11:18
well, you can't really convert 0 to an array – G_hi3 Aug 06 '15 at 12:28
he convert array to int it returns 0 – Marwan Aug 06 '15 at 14:28

why does "[] == 0" return true while "[]" is true and "0" is false?

2 Answers2