[Haskell-cafe] Existentially-quantified constructors, Eq and Show
John Meacham
john at repetae.net
Mon Dec 12 05:15:11 EST 2005
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⑈
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