Haskell Existential Types

Question

I'm trying to wrap my brain around Haskell's existential types, and my first example is a heterogeneous list of things that can be shown:

{-# LANGUAGE ExistentialQuantification #-}
data Showable = forall a. Show a => Showable a

showableList :: [Showable]
showableList = [Showable "frodo", Showable 1]

Now it seems to me that the next thing I would want to do is make Showable an instance of Show so that, for example, my showableList could be displayed in the repl:

instance Show Showable where
  show a = ...

The problem I am having is that what I really want to do here is call the a's underlying show implementation. But I'm having trouble referring to it:

instance Show Showable where
  show a = show a

picks out Showable's show method on the RHS which runs in circles. I tried auto-deriving Show, but that doesn't work:

data Showable = forall a. Show a => Showable a
  deriving Show

gives me:

Can't make a derived instance of `Show Showable':
  Constructor `Showable' does not have a Haskell-98 type
  Possible fix: use a standalone deriving declaration instead
In the data type declaration for `Showable'

I'm looking for someway to call the underlying Show::show implementation so that Showable does not have to reinvent the wheel.

Answer 1

instance Show Showable where
   show (Showable a) = show a

show a = show a doesn't work as you realized because it recurses infinitely. If we try this without existential types we can see the same problem and solution

data D = D Int
instance Show D where show a = show a -- obviously not going to work

instance Show D where show (D a) = "D " ++ (show a) -- we have to pull out the underlying value to work with it