On 10 March 2011 14:47, Daniel Fischer wrote:

If memory serves correctly, it's impredicative polymorphism.

Indeed. For example the following also doesn't type check in GHC-7: foo :: (forall s. ST s a) -> a foo st = ($) runST st Surprisingly the following does: foo :: (forall s. ST s a) -> a foo st = runST $ st Because GHC contains a special rule for infix $. Also see: http://article.gmane.org/gmane.comp.lang.haskell.glasgow.user/19152 Bas

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

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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On 10 March 2011 17:55, Bas van Dijk wrote:

On 10 March 2011 18:24, Yves Parès wrote:

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

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

As explained by the email from SPJ that I linked to, instantiating a type variable (like 'b') with a polymorphic type (like 'forall s. ST s a' ) is called impredicative polymorphism. Since GHC-7 this is not supported any more because it was to complicated.

AFAIK this decision was reversed because SPJ found a simple way to support them. Indeed, they work fine in 7.0.2 and generate warnings. Try it out: {{{ {-# LANGUAGE ImpredicativeTypes #-} module Impred where f :: Maybe (forall a. [a] -> [a]) -> Maybe ([Int], [Char]) f (Just g) = Just (g [3], g "hello") f Nothing = Nothing }}} Unfortunately, the latest user guide still reflects the old situation: http://www.haskell.org/ghc/docs/latest/html/users_guide/other-type-extension... Cheers, Max

On 11 March 2011 11:15, Max Bolingbroke wrote:

On 10 March 2011 17:55, Bas van Dijk wrote:

...

On 10 March 2011 18:24, Yves Parès wrote:

...

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

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

As explained by the email from SPJ that I linked to, instantiating a type variable (like 'b') with a polymorphic type (like 'forall s. ST s a' ) is called impredicative polymorphism. Since GHC-7 this is not supported any more because it was to complicated.

AFAIK this decision was reversed because SPJ found a simple way to support them. Indeed, they work fine in 7.0.2 and generate warnings. Try it out:

Great! Unfortunately foo still doesn't type check in 7.0.2: foo :: (forall s. ST s a) -> a foo st = ($) runST st For the same reason I still need this ugly hack in usb-safe: http://hackage.haskell.org/packages/archive/usb-safe/0.12/doc/html/src/Syste... Bas