[Haskell-beginners] IO action on a list of [IO a]

[ugo pozo]

Here is my try. It works for any monad (not just IO) and any foldable
(not just lists).

import Data.Foldable

myfor :: (Monad m, Foldable t) => (a -> m ()) -> t (m a) -> m ()
myfor f = foldlM (flip $ const (>>=f)) ()

--
~ ugo pozo
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On Sat, Oct 6, 2012 at 9:08 PM, Brent Yorgey <byorgey at seas.upenn.edu> wrote:
> On Sat, Oct 06, 2012 at 07:35:13PM +0200, Henk-Jan van Tuyl wrote:
>> On Sat, 06 Oct 2012 16:16:18 +0200, Manfred Lotz
>> <manfred.lotz at arcor.de> wrote:
>>
>> >myfor :: (a -> IO () ) -> [IO a] -> IO ()
>> >myfor _ [] = return ()
>> >myfor f (x:xs) = do
>> >  x' <- x
>> >  f x'
>> >  myfor f xs
>> >
>> >
>> >Is there a library function doing just this?
>>
>> You could use this:
>>   import Control.Monad
>>   myfor :: (a -> IO () ) -> [IO a] -> IO ()
>>   myfor f (x:xs) = mapM_ (liftM f) xs
>
> This should be
>
>   myfor f xs = mapM_ (>>= f) xs
>
> using (liftM f) will result in the right type but it does the wrong
> thing, as Manfred observed:
>
>   f :: a -> IO ()
>   liftM f :: IO a -> IO (IO a)
>
> So mapping liftM f over a list of IO actions results in a list of IO
> actions with no effects, whose results are the IO actions you really
> wanted.  Then mapM_ throws away those IO actions you really wanted,
> resulting in essentially (return () :: IO ()).
>
> -Brent
>
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