Well, it seems that you can't do exactly what you want. So, the simplest way to
do this would be not to make Foo a superclass for Bar:
class Bar a where
foo :: Foo a b => a -> b -> c
Then you would have to mention Foo everywhere.
If you really need, for some reason, to ensure that every Bar instance has a
corresponding Foo instance, you can do some oleging this way:
data Void b = Void
data FooEv a where FooEv :: Foo a b => Void b -> FooEv a
class Bar a where
barFoo :: FooEv a
bar :: Foo a b => a -> b -> c
Then, whenever you need Foo methods, you can do pattern-matching:
case barFoo :: FooEv a of
FooEv (Void :: Void b) -> …
Now some "b" is in scope, and there is an instance of Foo a b.
On Sep 28, 2012, at 8:36 PM, Francesco Mazzoli <f...@mazzo.li> wrote:
> I would expect this to work, maybe with some additional notation (a la
> ScopedTypeVariables)
>
> {-# LANGUAGE FunctionalDependencies #-}
> {-# LANGUAGE MultiParamTypeClasses #-}
>
> class Foo a b | a -> b
>
> class Foo a b => Bar a where
> foo :: a -> b -> c
>
> The type family equivalent works as expected:
>
> {-# LANGUAGE TypeFamilies #-}
>
> class Foo a where
> type T a :: *
>
> class Bar a where
> foo :: a -> T a -> c
>
> I can't use type families because the `Foo' I'm using is in an external
> library.
> Is there any way to achieve what I want without adding `b' to `Bar'?
>
> --
> Francesco * Often in error, never in doubt
>
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