Show instance of Void

[ersran9]

This is less of an issue and more of a query, but why does Void have a Show instance?
Doing a show on a value of Void goes into an infinite loop anyway.

[ekmett]

The Show instance exists because if you have something like [Void] you can then Show it.

You sould never construct a value of kind Void.

But you can easily have something like

data Exp a = Var a | App (Exp a) (Exp a) | Lam (Exp (Maybe a))
  deriving Show

and talk about closed terms as Exp Void and want to show them.