What is the kind of Void?

Question

Seeing as the type of Void is uninhabited, can it be regarded as a type \"constructor\"? Or is this just a quick \"hack\" to be able to safely disregard / disable f

Answer 1

Another angle on this question: suppose I asked you to write a guaranteed-terminating function of type a -> b:

aintGonnaWork :: a -> b
aintGonnaWork a = _

As hopefully you can tell, it's impossible to write such a function. It follows from this that the type a -> b has no defined values. Note also that the kind of a -> b is *:

(->)   :: * -> * -> *
a      :: *
b      :: *
---------------------
a -> b :: *

And there we have it: a type of kind *, built from "vanilla" Haskell elements (no "hacks"), but which nevertheless has no defined values. So the existence of types like Void is already implicit in "vanilla" Haskell; all that the explicit Void type does is provide a standard, named one.

I'll close with a simple implementation of the Void type in terms of the above; the only extension necessary is RankNTypes.

{-# LANGUAGE RankNTypes #-}

newtype Void = Void (forall a b. a -> b)

absurd :: Void -> a
absurd (Void f) = f f