[Haskell-beginners] suggestions for re-implementing mapM
David McBride
toad3k at gmail.com
Fri Mar 20 20:07:56 UTC 2015
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 at 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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>
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