[Haskell-cafe] Class constraints with "free" type variables and fundeps

[Francesco Mazzoli]

At Sat, 29 Sep 2012 13:04:59 +0400,
MigMit wrote:
> Well, it seems that you can't do exactly what you want. So, the simplest way
> to do this would be not to make Foo a superclass for Bar:
> 
> class Bar a where
>     foo :: Foo a b => a -> b -> c
> 
> Then you would have to mention Foo everywhere.

Just to clarify, I already worked my way around that problem.  I want to know
why I can't do it since it seems a useful feature to have.

In fact, I was doing something very similar to what you're proposing before, but
I think having the additional argument is better in my code, for various
reasons.  You can check the actual class here:
<https://github.com/bitonic/language-spelling/blob/c3b11111fa3014983acf41f9248c9507d7404424/Language/Distance/Search/Class.hs>,
I've left the old version commented.

> If you really need, for some reason, to ensure that every Bar instance has a
> corresponding Foo instance, you can do some oleging this way:
> 
> data Void b = Void
> data FooEv a where FooEv :: Foo a b => Void b -> FooEv a
> class Bar a where
>     barFoo :: FooEv a
>     bar :: Foo a b => a -> b -> c
> 
> Then, whenever you need Foo methods, you can do pattern-matching:
> 
> case barFoo :: FooEv a of
>   FooEv (Void :: Void b) -> …
> 
> Now some "b" is in scope, and there is an instance of Foo a b.

Well, I'm sure there are endless ways to get around this, but in cases like this
the resulting mess far outweighs the advantages :).

--
Francesco * Often in error, never in doubt