On Wed, Aug 17, 2011 at 4:49 PM, Tom Schouten
<tom@zwizwa.be> wrote:
{-# 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)
Not without moving the quantifier, as Oleg says. Here is why:
Existential types are equivalent to packing up a type with the constructor; imagine KI as
data KI i o = KI @s (i -> s -> (s,o)) -- not legal haskell
where @s represents "hold a type s which can be used in the other elements of the structure". An example element of this type:
KI @[Int] (\i s -> (i:s, sum s)) :: KI Int Int
Trying to implement iso as you suggest, we end up with
iso f = KI ?? (\i s -> case f i of ...)
What do we put in the ??. In fact, it's not possible to find a solution; consider this:
ki1 :: KI () Int
ki1 = KI @Int (\() s -> (s+1, s))
ki2 :: KI () Int
ki2 = KI @() (\() () -> ((), 0))
f :: Bool -> KI () Int
f x = if x then ki1 else ki2
iso f = KI ?? ??
The problem is that we have multiple possible internal state types!
-- ryan
{-
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)
-}
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