I think you'll find a problem using do-notation with your Monad.
Tony Morris
On 04/01/2011 11:33 PM, "Alexey Khudyakov"
I'd like to argue in opposition of making Functor a...
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.o...