Sometimes it is easier to write it with do notation and then rewrite it back in normal terms:

mapM :: Monad m => (a -> m b) -> [a] -> m [b]
mapM _ [] = return []
mapM f (x:xs) = do
  x' <- f x
  xs' <- mapM f xs
  return $ x':xs'

You can rewrite the second part step by step as:

mapM f (x:xs) = f x >>= \x' -> mapM f xs >>= \xs' -> return (x' : xs')

Also the base package does not use do notation.  It defines it much more elegantly:

mapM            :: Monad m => (a -> m b) -> [a] -> m [b]
mapM f as       =  sequence (map f as)

On Fri, Mar 20, 2015 at 3:52 PM, Vale Cofer-Shabica <vale.cofershabica@gmail.com> wrote:

Hello all,

I've been reading through "Tackling the Awkward Squad" [1] and am implementing the definitions "left as exercises" as I go. For section 2.2, I define:

    putLine :: [Char] -> IO ()
    putLine :: mapM_ putChar

where

    mapM_ :: Monad m => (a -> m b) -> a -> m ()
    mapM_ f [] = return ()
    mapM_ f (x:xs) = (f x) >> (mapM_ f xs)

which works without difficulty. For the sake of learning, I decided to implement mapM as well. My definition (below) works, but seems really in-elegant. I checked the prelude source and found mapM defined in terms of sequence, which has some do-notation I'm not so clear on. I'm trying to avoid using do notation in my implementation because it still feels like magic. I'll work on de-sugaring do notation again once I have a solid handle on (>>=), (>>), and return. Any suggestions for cleaning this up would be much appreciated!

    mapM :: Monad m => (a -> m b) -> [a] -> m [b]
    mapM f [] = return []
    mapM f (x:xs) = consMM (f x) (mapM f xs)

    consMM :: Monad m => m a -> m [a] -> m [a]
    consMM mx mxs = mx >>= ((flip consM)  mxs) where
        consM x mxs = mxs>>=(\xs -> return (x:xs))

Thank you,
vale

[1] Suggested by apfelmus in a recent email to the list:
http://research.microsoft.com/en-us/um/people/simonpj/papers/marktoberdorf/

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