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!

Answer 2

"Is false" (even with coercion) is different from "evaluates as false in boolean context." obj == false asks if the object is the boolean value false, not whether it would evaluate as such if evaluated in boolean context.

You can evaluate an object in boolean context with (!!obj).

[] == false;         // true
(!![]) == false;     // false

"\n  " == false;     // true
(!!"\n  ") == false; // false

Answer 3

from the ECMA-262 5.1 (page 83):

If ToBoolean(lval) is true, return lval

[] || false; // []