On Sun, Dec 11, 2005 at 07:41:43PM +0000, Ben Rudiak-Gould wrote:
I think the problem is not with the use of forall, but with the use of the term "existential type". The fact that existential quantification shows up in discussions of this language extension is a red herring. Even Haskell 98 has existential types in this sense, since (forall a. ([a] -> Int)) and ((exists a. [a]) -> Int) are isomorphic.
this is why exists makes more sense to me :) data Foo = exists a . Foo (a -> a) now if we simply deforest the following call, f :: Foo -> Int we get f :: (exists a . a -> a) -> Int which is the same as f :: forall a . a -> a -> Int which is the type we want. however, perhaps it is an argument for data Foo = Foo (exists a . a -> a) as the syntax. John -- John Meacham - ⑆repetae.net⑆john⑈