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Tue Aug 16 01:50:07 CEST 2011

## HC question: Hiding "growing" state using existentials.

See the thread at:
http://www.haskell.org/pipermail/haskell-cafe/2011-August/094673.html
Dear Cafe,
I'm building an abstraction for representing sequences as difference
equations, storing initial values and update equation.
I have something that resembles a Monad, but has an extra state
parameter s that "grows" on _join or _bind, so I can't simply create
an instance Monad (Sig s).
> {-# LANGUAGE ExistentialQuantification #-}
> data Sig s a = Sig s (s -> (a, s))
> _join :: (Sig s1 (Sig s2 a)) -> Sig (s1,s2) a
> _join (Sig i1 u1) = Sig (i1, i2) u12 where
> ((Sig i2 _), _) = u1 i1
> u12 (s1, s2) = (a, (s1', s2')) where
> ((Sig _ u2), s1') = u1 s1
> (a, s2') = u2 s2
> _fmap :: (a -> b) -> Sig s a -> Sig s b
> _fmap f (Sig s0 u) = Sig s0 u' where
> u' s = (f a, s') where
> (a, s') = u s
> _return x = Sig () $ \() -> (x, ())
> _bind :: (Sig s1 a) -> (a -> Sig s2 b) -> (Sig (s1,s2) b)
> m `_bind` g = _join ((_fmap g) m)
I try to hide the state parameter using existential quantification.
> data Signal a = forall s. Signal (Sig s a)
This approach works for defining Functor and Applicative instances,
but I can't seem to find a way to obtain an unwrapped version of f
that can be passed to _bind to implement the Monad instance's (>>=).
The following returns:
Couldn't match type `t' with `Sig s1 b'
`t' is a rigid type variable bound by
the inferred type of f' :: a -> t
at /home/tom/meta/ssm/SigHC.lhs:52:7
In the expression: mb
In a case alternative: Signal mb -> mb
In the expression: case (f a) of { Signal mb -> mb }
> instance Monad Signal where
> return = Signal . _return
> (>>=) (Signal ma) f = Signal $ ma `_bind` f' where
> f' a = case (f a) of
> Signal mb -> mb
Any tips on how to work around this?
Cheers,
Tom

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