Question

A Foldable instance is likely to be some sort of container, and so is likely to be a Functor as well. Indeed, this says

A Foldable type is also a container (although the class does not technically require Functor, interesting Foldables are all Functors).

So is there an example of a Foldable which is not naturally a Functor or a Traversable? (which perhaps the Haskell wiki page missed 🙂 )

Answer 1

Here’s a fully parametric example:

data Weird a = Weird a (a -> a)

instance Foldable Weird where
  foldMap f (Weird a b) = f $ b a

Weird is not a Functor because a occurs in a negative position.