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 <[email protected]> > 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 > > _______________________________________________ > Haskell-Cafe mailing list > [email protected] > http://www.haskell.org/mailman/listinfo/haskell-cafe >
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