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@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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