[Haskell-cafe] GADT and problems with rigid type variables
Daniel Peebles
pumpkingod at gmail.com
Sun Aug 22 18:47:36 EDT 2010
The problem is that you have an existential `t` there, and two values of the
type Foo might not have the same `t` inside them.
What do you want to happen if someone writes Foo True == Foo "yep"?
The only real solution here is to parametrize your Foo type by the t that
lives within it, so you can ensure the inner types are the same. You could
also do some (in my opinion) fairly nasty stuff with Dynamic or Typeable,
adding a constraint to the Eq and attempting to cast at runtime (returning
False if the cast returns Nothing).
Hope this helps!
On Mon, Aug 23, 2010 at 12:36 AM, Markus Barenhoff <alios at alios.org> wrote:
> Hello,
> playing with GADTs I ran into a problem with rigid type variables
> which is ilustrated by the following example. I think it should be
> pretty clear what I'am trying to express... Any suggestions?
>
> ---- snip ----
> {-# LANGUAGE GADTs #-}
>
> data Foo where
> Foo :: (Eq t) => t -> Foo
>
> instance Eq Foo where
> (Foo a) == (Foo b) = a == b
>
> {-
> Scratch.hs:7:28:
> Couldn't match expected type `t' against inferred type `t1'
> `t' is a rigid type variable bound by
> the constructor `Foo' at /home/alios/src/lab/Scratch.hs:7:3
> `t1' is a rigid type variable bound by
> the constructor `Foo' at /home/alios/src/lab/Scratch.hs:7:14
> In the second argument of `(==)', namely `b'
> In the expression: a == b
> In the definition of `==': (Foo a) == (Foo b) = a == b
> Failed, modules loaded: none.
> -}
> ---- snip ----
>
> thnx
> Markus
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