Re: Superclass Cycle via Associated Type

2011/7/22 Dan Doel :

2011/7/22 Gábor Lehel :

...

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 ... When I last tried this in the 7.0 time frame (before it was discovered that equality superclasses don't actually work yet and got disabled), it gave me a superclass cycle error for the first halves of those contexts, and for the second halves (the equality constraints) I believe it caused some kind of context stack depth to be exceeded (again because of the recursiveness). I'll have to try this again once I manage to get 7.2 compiled.

-- Dan

-- Work is punishment for failing to procrastinate effectively.