[Haskell-cafe] Can not use ST monad with polymorphic function
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
Re: [Haskell-cafe] Can not use ST monad with polymorphic function
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
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> Haskell-Cafe mailing list
> Haskell-Cafe@haskell.org
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Re: [Haskell-cafe] Can not use ST monad with polymorphic function
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
>
>
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Re: [Haskell-cafe] Can not use ST monad with polymorphic function
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.
--
Dave Menendez
Re: [Haskell-cafe] Can not use ST monad with polymorphic function
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
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Re: [Haskell-cafe] Can not use ST monad with polymorphic function
> > 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. > > Thank you, I have read about predicative types and it seems I understand the origin of the problem now. > GHC has an extension -XImpredicativeTypes that lifts this restriction, > but in my experience, it doesn't work very well. > Yes, it didn't help in my case. Thank you, Dmitry ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
