On 9/16/07, Mads Lindstrøm wrote:

Hi all

If I have this type:

data Foo a b = ...

and this class

class Bar (x :: * -> *) where ...

I can imagine two ways to make Foo an instance of Bar. Either I must "apply" the 'a' or the 'b' in (Foo a b). Otherwise it will not have the right kind. To "apply" the 'a' I can do:

instance Bar (Foo a) where ...

But what if I want to "apply" the 'b' ? How do I do that ?

Greetings,

Mads Lindstrøm

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(Redirecting to Haskell-Cafe) I think the easiest way to make it work is with a newtype wrapper around 'Foo a b' that swaps the arguments, like Brent mentioned. Without it you need some kind of type level lambda abstraction. Something like: instance Bar (\x -> Foo x b) where ... But this is not valid Haskell. cheers, Bas