[Haskell-cafe] fmap vs. liftM

[Miguel Mitrofanov]

> Problem is that from the idea Functor is a superclass of Monad, with  
> the
> property that "fmap == liftM".

[cut]

> The second relation can even not be expressed in Haskell 98.

Erm...

class Functor f where
     fmap :: (a -> b) -> f a -> f b
class Functor m => Monad m where
     return :: a -> m a
     join :: m (m a) -> m a
bind :: Monad m => m a -> (a -> m b) -> m b
bind mx f = join $ fmap f mx

Now liftM must be exactly equal to fmap.