[Haskell-cafe] Small displeasure with associated type synonyms

[Stefan Holdermans]

Dear all,

I was doing a little experimentation with associated type synonyms and  
ran into a small but unforeseen annoyance. The following, contrived,  
example snippets illustrate my pain:

First, let us declare some of the simplest classes exhibiting  
associated type synonyms that I can think of,

   class C a where
     type T a
     val :: T a

together with a trivial instance for nullary products:

   instance C () where
     type T () = ()
     val       = ()

But then, let us try an instance for binary products:

   instance (C a, C b) => C (a, b) where
     type T (a, b) = (T a, T b)
     val           = (val, val)

I really thought this would work out nicely, but GHC (version 6.8.2)  
gracefully gives me

   Couldn't match expected type `T a2' against inferred type `T a'
     Expected type: T (a2, b)
     Inferred type: (T a, T a1)
   In the expression: (val, val)
   In the definition of `val': val = (val, val)

   Couldn't match expected type `T b' against inferred type `T a1'
     Expected type: T (a2, b)
     Inferred type: (T a, T a1)
   In the expression: (val, val)
   In the definition of `val': val = (val, val)

while I think I deserve better than that.

Can someone (Tom?) please explain (a) why the required unifications  
fail, (b) whether or not it is reasonable to expect the unifications  
to succeed, and (c) how I can overcome problems like these? Surely, I  
can have val take a dummy argument, but I am hoping for something a  
bit more elegant here.

I tried lexically scoped type variables, but to no avail:

   instance forall a b. (C a, C b) => C (a, b) where
     type T (a, b) = (T a, T b)
     val           = (val :: T a, val :: T b)

gives me

   Couldn't match expected type `T a2' against inferred type `T a'
   In the expression: val :: T a
   In the expression: (val :: T a, val :: T b)
   In the definition of `val': val = (val :: T a, val :: T b)

etc.

Cheers,

   Stefan