[Haskell-cafe] ContT and ST stack

[Yves Parès]

Why has the operator (.) troubles with a type like (forall s. ST s a)?

Why can't it match the type 'b' in (.) definition?


2011/3/10 Daniel Fischer <daniel.is.fischer at googlemail.com>

> On Thursday 10 March 2011 14:18:24, Anakim Border wrote:
> > Dear list,
> >
> > I have the following (simplified) piece of code:
> >
> > find :: Int -> [Int]
> > find i = runST . (`runContT` return) $
> >   callCC $ \escape -> do
> >     return []
> >
> > which used to compile correctly under GHC 6.12.3.
> >
> > Now that I've switched to 7.0.2 it gets rejected with the following
> > error:
> >
> >     Couldn't match expected type `forall s. ST s c0'
> >                 with actual type `m0 r0'
> >     Expected type: ContT r0 m0 a0 -> forall s. ST s c0
> >       Actual type: ContT r0 m0 a0 -> m0 r0
> >     In the second argument of `(.)', namely `(`runContT` return)'
> >     In the expression: runST . (`runContT` return)
> >
> >
> > I'm a little bit lost at what exactly is the problem.
>
> If memory serves correctly, it's impredicative polymorphism.
>
> The type of (.), (b -> c) -> (a -> b) -> a -> c,  can't handle
> (x -> forall s. ST s [Int])
>
> Previously there was an implementation of impredicative polymorphism which
> allowed GHC to handle that construct, but that has been removed (because it
> was unsatisfactory), so GHC 7 doesn't compile that anymore.
>
> > Anyone can suggest a solution?
>
> Parentheses.
>
> find i = runST ((`runContT` return) $
>    callCC $ \escape -> do
>        return [])
>
> >
> > Thanks!
> >
> > AB
>
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