Type constraints on all type family instances

Question

I suppose what I want is impossible without Template Haskell but I'll ask anyway.

I have an interface for types like Data.Set and Data.IntSet:

type family Elem s :: *
class SetLike s where
  insert :: Elem s -> s -> s
  member :: Elem s -> s -> Bool
  ...

type instance Elem (Set a) = a
instance Ord a => SetLike (Set a) where
  ...

And I have a type family which chooses optimal set implementation:

type family EfficientSet elem :: *
type instance EfficientSet Int = IntSet
type instance EfficientSet String = Set String -- or another implementation

Is there a way to guarantee that EfficientSet instances will be always SetLike and that Elem (EfficientSet a) is a ?

Without this guarantee all function signatures will be like this:

type LocationSet = EfficientSet Location
f :: (SetLike LocationSet, Elem LocationSet ~ Location) => ...

To write each time SetLike LocationSet is somewhat tolerable, but Elem LocationSet ~ Location makes code understanding only harder, as for me.

Answer 1

Using GHC 7.4's constraint kinds you could have something like

type EfficientSetLike a = (SetLike (EfficientSet a),Elem (EfficientSet a) ~ a)

You can (with appropriate extensions) get constraints like this in earlier versions of GHC

class (SetLike (EfficientSet a),Elem (EfficientSet a) ~ a) => EfficientSetLike a 
instance (SetLike (EfficientSet a),Elem (EfficientSet a) ~ a) => EfficientSetLike a 

But, the new style type declaration is much nicer.

I'm not exactly sure what you are looking for, but it sounds like you just want easier to write/understand constraint signatures, in which case this will work.

Answer 2

You can write this:

class (SetLike (EfficientSet a), Elem (EfficientSet a) ~ a) =>
      HasEfficientSet a where
    type EfficientSet a

If you associate the Elem type family with the SetLike class, you probably wouldn't even need the SetLike superclass constraint:

class SetLike s where
    type Elem s
    insert :: Elem s -> s -> s
    member :: Elem s -> s -> Bool

class Elem (EfficientSet a) ~ a => HasEfficientSet a where
    type EfficientSet a

Answer 3

I like Daniel Wagner's post, but you can't just write:

test :: (HasEfficientSet a) => EfficientSet a
test = empty

You have to write:

test :: (HasEfficientSet a, SetLike (EfficientSet a)) => EfficientSet a
test = empty

But this can be overcome with a constraint synonym:

class SetLike s where 
    type Elem s :: *
    empty :: s
    insert :: Elem s -> s -> s 
    member :: Elem s -> s -> Bool 

class (Elem (EfficientSet a) ~ a) => HasEfficientSet' a where
    type EfficientSet a :: *

type HasEfficientSet a = (HasEfficientSet' a, SetLike (EfficientSet a))


newtype UnitSet = UnitSet Bool
    deriving (Show, Eq, Ord)

instance SetLike UnitSet where
    type Elem UnitSet = ()
    empty = UnitSet False 
    insert () _ = UnitSet True
    member () u = UnitSet True == u

instance HasEfficientSet' () where
    type EfficientSet () = UnitSet


test :: (HasEfficientSet a) => EfficientSet a
test = empty

test1 :: EfficientSet ()
test1 = empty

test2 :: EfficientSet ()
test2 = test

test3 :: UnitSet
test3 = empty

test4 :: UnitSet
test4 = test