[Haskell-beginners] yet another monad question

[Ovidiu Deac]

thanks for the explanations! I guess it's clear now

On Sat, Feb 4, 2012 at 3:02 PM, Chaddaï Fouché <chaddai.fouche at gmail.com>wrote:

> On Sat, Feb 4, 2012 at 12:05 PM, David McBride <toad3k at gmail.com> wrote:
> > When you pass an argument to
> > readDir = findRegularFiles >>= readFiles
> >
> > it expands to
> > readDir arg = (findRegularFiles >>= readFiles) arg
> >
> > which fails because that expression takes no argument, only
> > findRegularFiles does.  Honestly I can't think of any way to get that
> > argument in there without explicitly naming it.
>
> I would say the problem is even before that, the expression
> "findRegularFiles >>= readFiles" is not well typed :
>
> (>>=) :: Monad m => m a -> (a -> m b) -> m b
> specialized here in :
> (>>=) :: IO a -> (a -> IO b) -> IO b
>
> but :
>
> findRegularFiles :: FilePath -> IO [FilePath]
>
> so findRegularFiles is not of type "IO a", so can't be the first
> argument of (>>=) (or the second of (=<<) since that's just the
> flipped version).
>
> But there is a solution ! What you're searching here is a function of type
> :
> ? :: Monad m => (a -> m b) -> (b -> m c) -> (a -> m c)
>
> A kind of monadic composition, there is an operator for that in
> Control.Monad since ghc 6.12 or even before :
>
> (>=>) :: Monad m => (a -> m b) -> (b -> m c) -> (a -> m c)
>
> so :
>
> > readDir = findRegularFiles >=> readFiles
>
> or
>
> > readDir = readFiles <=< findRegularFiles
>
> will work :)
>
> --
> Jedaï
>
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