Re: Superclass Cycle via Associated Type

23 Jul 2011

      2011/7/23 Gábor Lehel 
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2011/7/22 Dan Doel :
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2011/7/22 Gábor Lehel :
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Yeah, this is pretty much what I ended up doing. As I said, I don't
think I lose anything in expressiveness by going the MPTC route, I
just think the two separate but linked classes way reads better. So
it's just a "would be nice" thing. Do recursive equality superclasses
make sense / would they be within the realm of the possible to
implement?
Those equality superclasses are not recursive in the same way, as far
as I can tell. The specifications for classes require that there is no
chain:
C ... => D ... => E ... => ... => C ...
However, your example just had (~) as a context for C, but C is not
required by (~). And the families involved make no reference to C,
either. A fully desugared version looks like:
type family Frozen a :: *
   type family Thawed a :: *
class (..., Thawed (Frozen t) ~ t) => Mutable t where ...
I think this will be handled if you use a version where equality
superclasses are allowed.
To be completely explicit, I had:
class (Immutable (Frozen t), Thawed (Frozen t) ~ t) => Mutable t where
type Frozen t ...
class (Mutable (Thawed t), Frozen (Thawed t) ~ t) => Immutable t where
type Thawed t ...
I had a similar issue in my representable-tries package.

In there I had

type family Key (f :: * -> *) :: *

class Indexable f where
  index :: f a -> Key f -> a

class Indexable f => Representable f where
  tabulate :: (Key f -> a) -> f a

such that tabulate and index witness the isomorphism from f a to (Key f ->
a).

So far no problem. But then to provide a Trie type that was transparent I
wanted.

class (Representable (BaseTrie e), Traversable (BaseTrie e), Key (BaseTrie
e) ~ e) => HasTrie e where
  type BaseTrie e :: * -> *

type (:->:) e = BaseTrie e

which I couldn't use prior to the new class constraints patch.

The reason I mention this is that my work around was to weaken matters a bit

class (Representable (BaseTrie e)) => HasTrie e where
  type BaseTrie e :: * -> *
  embedKey :: e -> Key (BaseTrie e)
  projectKey :: Key (BaseTrie e) -> e

This dodged the need for superclass equality constraints by just requiring
me to supply the two witnesses to the isomorphism between e and Key
(BaseTrie e).

Moreover, in my case it helped me produce instances, because the actual
signatures involved about 20 more superclasses, and this way I could make
new HasTrie instances for newtype wrappers just by defining an embedding and
projection pair for some type I'd already defined.

But, it did require me to pay for a newtype wrapper which managed the
embedding and projection pairs.

newtype e :->: a = Trie (BaseTrie e a)

In your setting, perhaps something like:

type family Frozen t
type family Thawed t

class Immutable (Frozen t) => Mutable t where
  thawedFrozen :: t -> Thawed (Frozen t)
  unthawedFrozen :: Thawed (Frozen t) -> t

class Mutable (Thawed t) => Immutable t where
  frozenThawed :: t -> Frozen (Thawed t)
  unfrozenThawed :: Frozen (Thawed t) -> t

would enable you to explicitly program with the two isomorphisms, while
avoiding superclass constraints.

-Edward

Re: Superclass Cycle via Associated Type

Edward Kmett