On 11/2/05, Jonathan Lang <[EMAIL PROTECTED]> wrote:
> Let's say you have this:
>
> role A {method foo() { code1; } }
> role B {method foo() { code2; } }
> role C does A does B {
> method foo() { A::foo(); }
> method bar() { B::foo(); }
> }
>
> Should the following be valid?
>
> role D does C { method foo() { B::foo(); } }
>
> IMHO, it shouldn't, because D doesn't do B.
Of course it should. To me, "role D does C" says, not "anything that
does D must do C", but "anything that does D does C".
It makes sense if you think of them as interfaces. Say you have a
role Complexifiable (that is, this thing can behave like a complex
number). Then you have:
role Numifiable does Complexifiable {
...
}
Complexifiable is just an abstraction that is used rarely; most things
will implement Numifiable. But if you can behave like a real number,
certainly you can behave like a complex number. But you don't want
people to have to say:
class Foo {
does Complexifiable; # wtf? what does that mean
does Numifiable;
...
}
Haskell makes a distinction between the two cases we're talking about here:
-- You need to be Complexifiable before you can be Numifiable
class (Complexifiable a) => Numifiable a where
...
-- Anything that is Numifiable is also Complexifiable
-- (by these rules):
instance Numifiable a => Complexifiable a where
...
To me, "does" is much more like the latter. Maybe we have constraints
on roles that can do the former[1]:
role Numifiable {
where Complexifiable;
...
}
Or maybe we don't, and if you need them to implement Complexifiable
first, you just leave some methods undefined that they have to define
themselves, thus completing the interface. I was always annoyed at
Haskell because I had to define an instance of Eq before I could
define an instance of Ord, even though Ord's algebra implies that I
can compare things for equality (and especially, since Ord defines a
total order, that a <= b and b <= a implies a = b, so it could have
implemented that for me).
It kind of seems like a Perlish thing to do to blur the two behaviors
together.
Luke
[1] Which is just this theory, without any extensions:
theory Numifiable{^T} <= Complexifiable{^T} {
...
}
So it would just be sugar in any case.
