Thank you, MigMit! If I replace your type FoldSTVoid with: data FoldMVoid = FoldMVoid {runFold :: Monad m => (Int -> m ()) -> m ()} then everything works magically with any monad! That is exactly what I wanted, though I still do not quite understand why wrapping the type solves the problem Dmitry On Thu, Nov 29, 2012 at 12:01 AM, MigMit wrote:

Yes, monomorphism. "do" binding requires your fold'' to be of some monomorphic type, but runST requires some polymorphism.

If you want, you can use special type like that:

data FoldSTVoid = FoldSTVoid {runFold :: forall a. (Int -> ST a ()) -> ST a ()}

fold :: Monad m => (Int -> m ()) -> m () fold f = mapM_ f [0..20]

selectFold :: String -> IO FoldSTVoid -- ((Int -> m ()) -> m ()) selectFold method = do -- in real program I'd like to choose between -- different fold methods, based on some IO context return $ FoldSTVoid fold

useFold :: FoldSTVoid -> ST a () useFold fold' = runFold fold' f where f _ = return () -- some trivial iterator

main = do fold'' <- selectFold "some-method-id" print $ runST $ useFold fold''

On Nov 28, 2012, at 9:52 PM, Dmitry Kulagin wrote:

...

Hi Cafe,

I try to implement some sort of monadic fold, where traversing is polymorphic over monad type. The problem is that the code below does not compile. It works with any monad except for ST. I suspect that monomorphism is at work here, but it is unclear for me how to change the code to make it work with ST.

fold :: Monad m => (Int -> m ()) -> m () fold f = mapM_ f [0..20]

selectFold :: Monad m => String -> IO ((Int -> m ()) -> m ()) selectFold method = do -- in real program I'd like to choose between -- different fold methods, based on some IO context return fold

useFold :: Monad m => ((Int -> m ()) -> m ()) -> m () useFold fold' = fold' f where f _ = return () -- some trivial iterator

main = do fold'' <- selectFold "some-method-id" print $ runST $ useFold fold''

Thank you! Dmitry _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

David Menendez wrote:

On Thu, Nov 29, 2012 at 7:50 AM, Dmitry Kulagin wrote:

...

Thank you, MigMit!

If I replace your type FoldSTVoid with: data FoldMVoid = FoldMVoid {runFold :: Monad m => (Int -> m ()) -> m ()}

then everything works magically with any monad! That is exactly what I wanted, though I still do not quite understand why wrapping the type solves the problem

Short answer: It's because GHC's type system is predicative.

Basically, quantified types can't be given as arguments to type constructors (other than ->, which is its own thing). I'm not entirely sure why, but it apparently makes the type system very complicated from a theoretical standpoint. By wrapping the quantified type in a newtype, the argument to IO becomes simple enough not to cause problems.

GHC has an extension -XImpredicativeTypes that lifts this restriction, but in my experience, it doesn't work very well. A newtype data Foo = Foo { bar :: forall a . baz a } usually works best. Best regards, Heinrich Apfelmus -- http://apfelmus.nfshost.com