If ([] == false) is true, why does ([] || true) result in []?

Question

Was just doing some testing and I find this odd:

[] == false

Gives true, this makes sense because double equal only compares contents and not t

Answer 1

Type conversion is not related to falsy and truthy values.

What is truthy and what is falsy is defined by the ToBoolean function defined in the specs and [] is indeed truthy.

On the other hand, [] == false returns true because of the type conversion that happens during the evaluation of the expression.

The rules of type conversion say that for x == y

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

ToNumber results in 0 for false so we're left with the evaluation of [] == 0. According to the same rules

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

ToPrimitive results in an empty string. Now we have "" == 0. Back to our type conversion rules

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

ToNumber results in 0 for "" so the final evaluation is 0 == 0 and that is true!