(I've moved this from haskell' to haskell-cafe, because it's more a question about associated types than about resolving the MPTC dilemma.) Bulat Ziganshin writes:
Hello Manuel,
MMTC> My statement remains: Why use a relational notation if you can MMTC> have a functional one?
how about these examples?
MMTC> class Monad m => RefMonad m where MMTC> type Ref m :: * -> *
can i use `Ref` as type function? for example:
data StrBuffer m = StrBuffer (Ref m Int) (Ref m String)
This is something I've been wondering about for a while. Can you do that
sort of thing with associated types?
As another example, consider this type class for composible continuation
monads:
class Monad m => MonadCC p sk m | m -> p sk where
newPrompt :: m (p a)
pushPrompt :: p a -> m a -> m a
withSubCont :: p b -> (sk a b -> m b) -> m a
pushSubCont :: sk a b -> m a -> m b
You can use instances of this class to create backtracking monads, along
these lines:
data Tree m a = HZero | HOne a | HChoice a (m (Tree m a))
newtype SR p m a = SR (forall ans. ReaderT (p (Tree m ans)) m a)
instance MonadCC p sk m => MonadPlus (SR p m)
With associated types, the MonadCC class becomes:
class Monad m => MonadCC m where
type Prompt m a
type SubCont m a b
newPrompt :: m (Prompt m a)
pushPrompt :: Prompt m a -> m a -> m a
withSubCont :: Prompt m b -> (SubCont m a b -> m b) -> m a
pushSubCont :: SubCont m a b -> m a -> m b
Since |Prompt m| is determined by |m|, can we eliminate the prompt
parameter from SR?
newtype SR' m a
= SR' (forall ans. ReaderT (Prompt m (Tree m ans)) m a)
instance MonadCC m => MonadPlus (SR' m)
That would presumably lead to SR' having the type
SR' :: (MonadCC m) => ReaderT (Prompt m (Tree m ans)) m a -> SR' m a
That doesn't seem like it should be a problem, since it's impossible to
create a value of type |ReaderT (Prompt m (Tree m ans)) m a| unless m is
an instance of MonadCC, but it also puts a context on a newtype data
constructor, which isn't currently allowed.
Sure, it's still possible to do this:
instance MonadCC m => MonadPlus (SR (Prompt m) m)
but that doesn't feel like a big win.
--
David Menendez