From what I know, the if statement in JavaScript casts the result of its condition to a Boolean, and then executes it like the following:
if(true) {
// run this
}
if(false) {
// do not run this
}
And that works. But If I do this:
if('0' == false) {
// We get here, so '0' is a falsy value
}
Then I would expect this:
if('0') {
// We don't get here, because '0' is falsy value
}
But instead, I get:
if('0') {
// We *DO* get here, even though '0' is falsy value
}
What's happening? Apparently, if does not check if its condition is a truthy or falsy value, but does some other conversion?
This is just one of those "gotchas" with the == rules which are rather complex.
The comparison x == y, where x and y are values, produces true or false. Such a comparison is performed as follows:
(4) If Type(x) is Number and Type(y) is String, return the result of the comparison x == ToNumber(y).
(5) If Type(x) is String and Type(y) is Number, return the result of the comparison ToNumber(x) == y.
(6) If Type(x) is Boolean, return the result of the comparison ToNumber(x) == y.
(7) If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).
In this case, that means that '0' == false is first coerced to '0' == 0 (by rule #7) and then, on the second pass through, it is coerced to 0 == 0 (by rule #5) which results in true.
This particular case is somewhat tricky because of false ~> 0 instead of '0' ~> true (as what might be expected). However, '0' is itself a truth-y value and the behavior can be explained with the above rules. To have strict truthy-falsey equality in the test (which is different than a strict-equality) without implicit conversions during the equality, consider:
!!'0' == !!false
(For all values: !falsey -> true and !truthy -> false.)