Empty arrays seem to equal true and false at the same time

Answer 1

Regarding the line:

if (arr == false) console.log("It's false!");

Maybe these will help:

console.log(0 == false) // true
console.log([] == 0) // true
console.log([] == "") // true

What I believe is happening is that the boolean false is coerced to 0 for comparison with an object (the left-hand side). The object is coerced to a string (the empty string). Then, the empty string is coerced into a number, as well, namely zero. And so the final comparison is 0 == 0, which is true.

Edit: See this section of the spec for details on exactly how this works.

Here's what's happening, starting at rule #1:

1. If Type(x) is different from Type(y), go to step 14.

The next rule that applies is #19:

19. If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).

The result of ToNumber(false) is 0, so we now have:

[] == 0

Again, rule #1 tells us to jump to step #14, but the next step that actually applies is #21:

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

The result of ToPrimitive([]) is the empty string, so we now have:

"" == 0

Again, rule #1 tells us to jump to step #14, but the next step that actually applies is #17:

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

The result of ToNumber("") is 0, which leaves us with:

0 == 0

Now, both values have the same type, so the steps continue from #1 until #7, which says:

7. If x is the same number value as y, return true.

So, we return true.

In brief:

ToNumber(ToPrimitive([])) == ToNumber(false)