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 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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe