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 <polyom...@f2s.com> wrote:
> Ryan Ingram <ryani.spam <at> gmail.com> 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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