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

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 ~ mailto:ugopozo@gmail.com ~ home://11.3266.2021 ~ cell://11.8432.9252 On Sat, Oct 6, 2012 at 9:08 PM, Brent Yorgey 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 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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