On 04.01.2011 13:24, o...@okmij.org wrote:
I'd like to argue in opposition of making Functor a super-class of Monad. I would argue that superclass constraints are not the right tool for expressing mathematical relationship such that all monads are functors and applicatives. Then argument is practical. It seems that making Functor a superclass of Monad makes defining new monad instances more of a chore, leading to code duplication. To me, code duplication is a sign that an abstraction is missing or misused.
I think I understood your point. But it looks like that it's possible to use subclass's function in superclass instance. At very least GHC is able to do it.
Following example works just fine without any language extensions in GHC6.12.3
import Prelude hiding (Monad(..), Functor(..)) class Functor f where fmap :: (a -> b) -> f a -> f b class Functor m => Monad m where return :: a -> m a (>>=) :: m a -> (a -> m b) -> m b instance Functor Maybe where fmap f m = m >>= (return . f) instance Monad Maybe where return = Just Nothing >>= _ = Nothing Just x >>= f = f x _______________________________________________ Haskell-prime mailing list Haskell-prime@haskell.org http://www.haskell.org/mailman/listinfo/haskell-prime
