[Haskell-cafe] Deconstruction

Hi, while this works: 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