An example of a Foldable which is not a Functor (or not Traversable)?

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<

Answer 1

Reading Beautiful folding I realized that any Foldable can be made a Functor by wrapping it into

data Store f a b = Store (f a) (a -> b)

with a simple smart contructor:

store :: f a -> Store f a a
store x = Store x id

(This is just a variant of the Store comonad data type.)

Now we can define

instance Functor (Store f a) where
    fmap f (Store x g)   = Store x (f . g)

instance (F.Foldable f) => F.Foldable (Store f a) where
    foldr f z (Store x g)    = F.foldr (f . g) z x

This way, we can make both Data.Set.Set and Sjoerd Visscher's Weird a functor. (However, since the structure doesn't memoize its values, repeatedly folding over it could be very inefficient, if the function that we used in fmap is complex.)

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Update: This also provides an example of a structure that is a functor, foldable but not traversable. To make Store traversable, we would need to make (->) r traversable. So we'd need to implement

sequenceA :: Applicative f => (r -> (f a)) -> f (r -> a)

Let's take Either b for f. Then we'd need to implement

sequenceA' :: (r -> Either b a) -> Either b (r -> a)

Clearly, there is no such function (you can verify with Djinn). So we can neither realize sequenceA.