On Sun, Jan 18, 2009 at 3:23 AM, Andrew Coppin
Given that liftM exists, why is having an identical implementation for fmap useful?
For many structures, it's easier to define (>>=) in terms of fmap and join. For these objects, often the "generic" implementation of liftM is far less efficient than fmap. That is to say, given a monad T and these functions: returnT :: a -> T a fmapT :: (a -> b) -> T a -> T b joinT :: T (T a) -> T a We can create Haskell instances as follows instance Functor T where fmap = fmapT instance Monad T where return = returnT m >>= f = joinT (fmap f m) Then, liftM f m = m >>= \x -> return (f x) = joinT (fmapT (\x -> return (f x)) m) Now, we know (by the monad & functor laws) that this is equivalent to (fmap f m), but it's a lot harder for the compiler to spot that. The list monad is a great example; I'd expect that using fmap (== map) in the list monad is significantly more efficient than liftM which constructs a singleton list for each element of the input and concatenates them all together. -- ryan