Re: [Haskell-beginners] yet another monad question

thanks for the explanations! I guess it's clear now On Sat, Feb 4, 2012 at 3:02 PM, Chaddaï Fouché wrote:

On Sat, Feb 4, 2012 at 12:05 PM, David McBride 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