In opposition of Functor as super-class of Monad
tmorris at tmorris.net
Tue Jan 4 14:38:33 CET 2011
I think you'll find a problem using do-notation with your Monad.
On 04/01/2011 11:33 PM, "Alexey Khudyakov" <alexey.skladnoy at gmail.com>
On 04.01.2011 13:24, oleg at okmij.org wrote:
> 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
Following example works just fine without any language extensions in
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
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