Re: [Haskell-cafe] fmap vs. liftM

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

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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.