On Tue, Nov 8, 2011 at 11:49 AM, Anupam Jain wrote:

While I understand why I get this error, I have no idea how to fix it! I cannot put a Show constraint on o1 because that variable is not exposed in the type of the expression.

That means 'o1' is an existencial variable.

I can work around this by changing my data type declaration to include Show constraints but I don't want to restrict my data type to only Showable things just so I could have a "Show" instance for debugging -

Only ∷ Show o ⇒ o → T o TT ∷ (Show o1, Show o2) ⇒ T o1 → (o1 → o2) → T o2

What else can I do to declare a Show instance for my datatype?

That's the only way of showing o1. Note that you don't need 'Show o2' constraint on 'TT', so this would suffice: data T o where Only :: o -> T o TT :: Show o1 => T o1 -> (o1 -> o2) -> T o2 instance Show o => Show (T o) where ... Without the inner Show constraint on TT you can't do any showing with o1. Since it is an existencial, it could be anything, even things that don't have Show instances. I think you may do something more complicated with the new ConstraintKinds extesions, something like data T c o where Only :: o -> T o TT :: c o1 => T o1 -> (o1 -> o2) -> T o2 instance Show o => Show (T Show o) where ... This is completely untested. And even if it works, I don't know if it is useful =). HTH, -- Felipe.

Hi, Anupam Jain wrote:

-- My datatype data T o where Only ∷ o → T o TT ∷ T o1 → (o1 → o2) → T o2

-- Show instance for debugging instance Show o ⇒ Show (T o) where show (Only o) = "Only " ⊕ (show o) show (TT t1 f) = "TT (" ⊕ (show t1) ⊕ ")"

As you noticed, the last line doesn't work because t1 is of some unknown type o1, and we know nothing about o1. In particular, we don't know how to show values of type t1, neither at compile time nor at runtime. I can't see a way around that. However, for debugging, you could do: show (TT t1 f) = "TT (" ++ show (f t1) ++ ")" This is not perfect, but at least it should work. Tillmann