Why is `[] == false` is true but just `[]` evaluates to true? [duplicate]

Answer 1

[] is an empty array and different than false that is why [] == false returns false WHILE [] returns true because it is not pointing on a null

Answer 2

The == operator forces a type cast. [] as boolean is false. Why doesn't the same happen with eg. "a" == false? Basically, the numeric value of [] is 0, while the numeric value of "a" is NaN. Predictably, "0" == false is true :)

On the other hand, in the second case, you're basically just checking if [] exists ("isn't null"). There is no casting to boolean.

Answer 3

Why is [] == false is true

Because arrays behave oddly when compared to primitive values.

In particular, when you compare any non-boolean to a boolean, the boolean is handled as a number. Then, when you compare a number to an object the object is converted to a primitive - which stringifies the array before again comparing it to the number. Now, that string is converted to a number so that they can be compared:

[] == false
[] == 0
"" == 0
0 == 0

Similarly, you can try

[1] == true
[1] == 1
"1" == 1
1 == 1

or

[2] == true
[2] == 1
"2" == 1
2 == 1

but just [] evaluates to true?

Because any object is truthy.

Answer 4

The first test returns true because of these equality rules :

If one of the operands is Boolean, the Boolean operand is converted to 1 if it is true and +0 if it is false.

If an object is compared with a number or string, JavaScript attempts to return the default value for the object. Operators attempt to convert the object to a primitive value, a String or Number value, using the valueOf and toString methods of the objects. If this attempt to convert the object fails, a runtime error is generated.

(and yes, you do have [].toString()=="" and thus []==0)

The second test

if ([])

simply is passed because all objects are truish.