On Oct 19, 2014, at 00:56 , Rick Mann <rm...@latencyzero.com> wrote:
>
> The subclass initializer still has to initialize itself. It knows what the
> superclass initializer is doing, and it knows what it still needs to do.
> That's true even with the rules Swift currently imposes.
I don’t actually know what “initialize itself” means. Nor do I see how it
“knows” what the superclass is doing, or at least I’m uncertain how this helps.
>> In Swift, an external caller cannot initialize a class by calling one of its
>> superclass initializers
> Of course, not, and I'm not suggesting that at all.
FWIW, I’m not sure there’s an “of course” here. In Obj-C, of course it can
literally do this.
However, going back to your original post, you wrote this:
> I tried this:
>
> 16 override
> 17 init()
> 18 {
> 19 init(windowNibName: "foo", owner: self);
> 20 }
Re-reading the Swift book to research my answers in this thread, I found this:
> “If your subclass provides an implementation of all of its
> superclass-designated initializers—either by inheriting them as per rule 1,
> or […] — then it automatically inherits all of the superclass convenience
> initializers.”
According to that, your original attempt should have worked, but it didn’t
because of a wrinkle:
> “[I]f you write a subclass initializer that matches a superclass convenience
> initializer, that superclass convenience initializer can never be called
> directly by your subclass […]. Therefore, you don not write the override
> modifier when providing a matching implementation of a superclass convenience
> initializer.”
In short, take the “override” keyword off your declaration, and it should work
exactly as you expected. (!)
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