data Foo a = Foo a

unwrapFoo :: Foo a -> a

unwrapFoo (Foo x) = x

this:

{-# LANGUAGE ExistentialQuantification #-}

class BarLike a where

doSomething :: a -> Double

data Bar = forall a. BarLike a => Bar a

unwrapBar :: Bar -> a

unwrapBar (Bar x) = x

gives me:

Couldn't match expected type `a' against inferred type `a1'

`a' is a rigid type variable bound by

the type signature for `unwrapBar' at test.hs:8:20

`a1' is a rigid type variable bound by

the constructor `Bar' at test.hs:9:11

In the expression: x

In the definition of `unwrapBar': unwrapBar (Bar x) = x

How can i deconstruct the enclosed value of type a?

Thanks,

Lenny