[Haskell-cafe] how to Enumerate without lists?

Ben Doyle benjamin.peter.doyle at gmail.com
Wed Sep 5 15:37:07 UTC 2018

Try this:

data Enumerator a b where
  Enumerator :: a -> a -> Enumerator a a

instance Enum a => Foldable (Enumerator a) where
  foldMap f (Enumerator x y)
    | fromEnum x > fromEnum y = mempty
    | otherwise                           = f x <> foldMap f (Enumerator
(succ x) y)

Here we're using a GADT to express that our two-parameter Enumerator type
in practice always has a == b (at the type level).
Which lets us constrain the values inside our new Foldable structure, while
still having a type of kind (* -> *) like the the
typeclass requires.

On Wed, Sep 5, 2018 at 6:56 AM Johannes Waldmann <
johannes.waldmann at htwk-leipzig.de> wrote:

> Hi David,
> Thanks for responding.
> Let me re-phrase the technical question: in some hypothetical
> >       instance Foldable Enumerator where ...
> the methods (e.g., foldMap) would be overconstrained.
> Is there a way to still write something like it?
> It seems not, as shown by these examples:
> Data.EnumSet cannot implement Foldable because of  Enum k.
> http://hackage.haskell.org/package/enummapset/docs/Data-EnumSet.html
> Data.IntSet cannot implement Foldable because of   k ~ Int.
> - J.
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