[Haskell-beginners] suggestions for re-implementing mapM

[Vale Cofer-Shabica]

Thanks, David

I appreciate the suggestions.

vale


On Fri, Mar 20, 2015 at 4:07 PM, David McBride <toad3k at gmail.com> wrote:

> 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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