It's hard. Here's a simple example:
type Foo f = f Int
class C (f :: (* -> *) -> *) where
thingy :: f [] -> f IO
-- Should this ever typecheck? I would say no; there's no way to
unify f [] with [Int].
callThingy :: [Int] -> IO Int
callThingy = thingy
-- but what if you say this?
instance C Foo where
-- thingy :: Foo [] -> Foo IO
-- therefore,
-- thingy :: [Int] -> IO Int
thingy (x:_) = return x
Basically, allowing instances for type functions requires you to
*un-apply* any type functions to attempt to do instance selection.
-- ryan
On Wed, Sep 29, 2010 at 2:15 PM, DavidA
Ryan Ingram
writes: Haskell doesn't have true type functions; what you are really saying is
instance Monad (\v -> Vect k (Monomial v))
Yes, that is exactly what I am trying to say. And since I'm not allowed to say it like that, I was trying to say it using a type synonym parameterised over v instead. It appears that GHC won't let you use partially applied type synonyms as type constructors for instance declarations. Is this simply because the GHC developers didn't think anyone would want to, or is there some theoretical reason why it's hard, or a bad idea?
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