On 9/16/07, Mads Lindstrøm <[EMAIL PROTECTED]> 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 > > > _______________________________________________ > Haskell mailing list > [EMAIL PROTECTED] > http://www.haskell.org/mailman/listinfo/haskell >
(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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
