At 2002-02-19 09:21, Simon Peyton-Jones wrote: >I don't know if it makes sense. You've written down some syntax, >but it's not clear to me what you intend by it.
Hmm... it should be straightforward... > instance (forall a. Eq a => Eq (f a)) => Eq (Rose f a) where.. I assume that the 'a' quantified in (forall a. ...) is not the same as the 'a' in 'Rose f a'? >| instance >| ( >| forall a. HasIdentity (m a a), >| forall a b c. Composable (m b c) (m a b) (m a c) >| ) => >| Category m; This means 'if for all a, "HasIdentity (m a a)", and also for all a b c, "Composable (m b c) (m a b) (m a c)", then "Category m"'. "(forall a. HasIdentity (m a a))" as a class assertion declares a property of m. It says that for all types a, there's an instance "HasIdentity (m a a)". >| Or even allow the foralls their own context: >| >| foo :: (forall a. (C a b) => D a c) => T b c; This means foo has type (T b c), where for every type a for which there's an instance "C a b", there's an instance "D a c". >| class >| ( >| forall a. HasIdentity (m a a), >| forall a b c. Composable (m b c) (m a b) (m a c) >| ) => >| Category m; Personally I think the 'superclass' arrow in class declarations should point the other way, as in '<='. After all, instances of the class imply instances of the superclasses, not the other way around. But that aside, this means '"Category" is a class on m, provided that for all types a, there's an instance "HasIdentity (m a a)", and also for all types a b c, there's an instance "Composable (m b c) (m a b) (m a c)"'. -- Ashley Yakeley, Seattle WA _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
