[Haskell-cafe] fmap vs. liftM
lemming at henning-thielemann.de
Tue Feb 5 08:31:57 EST 2008
On Mon, 4 Feb 2008, Miguel Mitrofanov wrote:
> > Problem is that from the idea Functor is a superclass of Monad, with
> > the
> > property that "fmap == liftM".
> > The second relation can even not be expressed in Haskell 98.
> 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.
How do you convince the compiler that
'join (fmap return x) == x' ?
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