GADTs and functional dependencies

[Jason Dagit]

On Tue, Sep 23, 2008 at 9:36 AM, Wolfgang Jeltsch
<g9ks157k at acme.softbase.org> wrote:
> Am Dienstag, 23. September 2008 18:19 schrieben Sie:
>> On Tue, Sep 23, 2008 at 6:07 PM, Wolfgang Jeltsch
>>
>> <g9ks157k at acme.softbase.org> wrote:
>> > Hello,
>> >
>> > please consider the following code:
>> >> {-# LANGUAGE GADTs, MultiParamTypeClasses, FunctionalDependencies #-}
>> >>
>> >> data GADT a where
>> >>
>> >>     GADT :: GADT ()
>> >>
>> >> class Class a b | a -> b
>> >>
>> >> instance Class () ()
>> >>
>> >> fun :: (Class a b) => GADT a -> b
>> >> fun GADT = ()
>> >
>> > I'd expect this to work but unfortunately, using GHC 6.8.2, it fails with
>> > the following message:
>>
>> bear in mind that the only instance you defined is
>>
>> instance Class () ()
>>
>> which doesn't involve your GADT at all.
>
> This is correct.  (It's only a trimmed-down example, after all.)
>
>> Maybe you meant something like:
>>
>> instance Class (GADT a) ()
>
> No, I didn't.
>
>> Moreover, your fun cannot typecheck, regardless of using MPTC or
>> GADTs. The unit constructor, (), has type () and not b.
>
> Pattern matching against the data constructor GADT specializes a to ().  Since
> Class uses a functional dependency, it is clear that b has to be ().  So it
> should typecheck.  At least, I want it to.  ;-)

In the signature:
fun :: (Class a b) => GADT a -> b

What is there to determine the type a?  It's a phantom in the
definition of GADT, so it can unify arbitrarily for us, but there is
nothing in your program to cause it to unify a certain way.

Based on what you're hoping for, I think you would need:
class Class a b | b -> a

But, then think about how the type checker looks at fun, which returns
().  Now we see that b should be (), but () is not the same as b.
That is, () does not generalize to b, even though an arbitrary b could
specialize to ().

Did some other version of ghc accept this code?

Jason