Hello,
On Mon, 21 Mar 2005 00:29:38 +0100, Thomas Jäger
Hello,
Aside from naming issues, there seem to be some problems with the way FunctorM is currently implemented.
First of all, `FunctorM` should be a superclass of `Functor' because there is an obvios implementation of fmap in terms of fmapM
import Data.FunctorM import Control.Monad.Identity
fmap' :: FunctorM f => (a -> b) -> f a -> f b fmap' f = runIdentity . fmapM (return . f) It is already annyoing enough that `Funtor' isn't a subclass of `Monad' although every monad must also be functor.
I think you are right. Does anyone remember why "Functor" is not a superclass of "Monad"?
Now, FunctorM should be based on the simplest operations possible, which in this case is the distributive law and not a monadic version of fmap (which might be provided for efficiency reasons).
class Functor f => FunctorM' f where dist' :: Monad m => f (m a) -> m (f a) fmapM' :: Monad m => (a -> m b) -> f a -> m (f b)
dist' = fmapM' id The "id" should be "return".
fmapM' f = dist' . fmap f
-- for example instance FunctorM' [] where dist' = sequence fmapM' = mapM
Well this is what I thought at first as well, but then I was looking through my code and realized that I hardly ever use "sequence" but I use "mapM" a lot. But of course one could have both operations in the class (as in your example). -Iavor