[Haskell-cafe] Existential question

[Tom Schouten]

{-# LANGUAGE ExistentialQuantification #-}

-- Dear Cafe, this one works.
data Kl' s i o = Kl' (i -> s -> (s, o))
iso' :: (i -> Kl' s () o) -> Kl' s i o
iso' f = Kl' $ \i s -> (\(Kl' kl') -> kl' () s) (f i)

-- Is there a way to make this one work also?
data Kl i o = forall s. Kl (i -> s -> (s, o))
iso :: (i -> Kl () o) -> Kl i o
iso f = Kl $ \i s -> (\(Kl kl) -> kl () s) (f i)

{-
     Couldn't match type `s0' with `s'
       because type variable `s' would escape its scope
     This (rigid, skolem) type variable is bound by
       a pattern with constructor
         Kl :: forall i o s. (i -> s -> (s, o)) -> Kl i o,
       in a lambda abstraction
     The following variables have types that mention s0
       s :: s0 (bound at /home/tom/meta/haskell/iso.hs:11:17)
     In the pattern: Kl kl
     In the expression: \ (Kl kl) -> kl () s
     In the expression: (\ (Kl kl) -> kl () s) (f i)
-}