Thanks, David
I appreciate the suggestions.
vale
On Fri, Mar 20, 2015 at 4:07 PM, David McBride
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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