the MPTC Dilemma (please solve)

[Manuel M T Chakravarty]

Bulat Ziganshin:
> Sunday, March 19, 2006, 7:25:44 PM, you wrote:
> 
> >> i had a class which defines "default" reference type for monads:
> >> 
> >> class Ref m r | m->r where
> 
> to be exact,
> 
> class Ref m r | m->r, r->m where
> >>   newRef :: a -> m (r a)
> >>   readRef :: r a -> m a
> >>   writeRef :: r a -> a -> m ()
> 
> or even worser:
> 
> class Ref2 m r a | m a->r, r->m
> instance (Unboxed a) => Ref2 IO IOURef a
> instance (!Unboxed a) => Ref2 IO IORef a
> instance (Unboxed a) => Ref2 (ST s) (STURef s) a
> instance (!Unboxed a) => Ref2 (ST s) (STRef s) a
> 
> MMTC> My statement remains:  Why use a relational notation if you can have a
> MMTC> functional one?
> 
> how about these examples?

Bidirectional FDs correspond to associated data types.  I am not sure
what you try to achieve with your Ref2 example.  The dependency of r on
a seems to be surious as you don't instantiate a in any of your instance
declarations.

> 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)

Yes, absolutely.

Manuel