deriving over renamed types
Mon, 8 Apr 2002 15:14:54 -0700
At 2002-04-08 12:45, Lennart Augustsson wrote:
>I just just wanted to say that I agree with almost everything Conor said.
>I find it a little odd that the extension to Haskell that allows explicit
>does not also allow you use explicit type application (and type lanbda).
What did you have in mind?
data Succ n;
type Add Zero b = b;
type Add (Succ a) b = Succ (Add a b);
type Mult Zero b = Zero;
type Mult (Succ a) b = Add b (Mult a b);
type Fact Zero = Zero;
type Fact (Succ n) = Mult (Succ n) (Fact n);
data Foo f = MkFoo (f ());
type Succ' = Succ;
type Succ'' n = Succ n;
-- which of these types are the same?
f1 = MkFoo undefined :: Foo Succ;
f2 = MkFoo undefined :: Foo Succ';
f3 = MkFoo undefined :: Foo Succ'';
f4 = MkFoo undefined :: (Add (Succ Zero));
Ashley Yakeley, Seattle WA