[Haskell] Re: "exists" keyword and "existential" types
ashley at semantic.org
Mon Feb 16 02:56:34 EST 2004
In article <Pine.LNX.4.21.0402150140100.16014-100000 at dark.darkweb.com>,
Ben Rudiak-Gould <benrg at dark.darkweb.com> wrote:
> And it would be nice to be able to pass around values of type
> (exists t. Interface t => t), which behave just like OOP interface
A value of "type" (exists t. Interface t => t) consists of two values,
one of type t, and one "dictionary" value. For that reason a data type
is used to represent this (and a newtype type cannot be).
So what's the difference? Data provides another layer of "thunkage". For
data D = MkD Int;
newtype N = MkN Int;
Then (MkN undefined) is the same as undefined, but (MkD undefined) is
So how does this apply to (exists t. Interface t => t)? Well, you'd have
two different versions of "undefined" depending on whether calculation
of the dictionary was part of the undefinition.
> I don't see how
> openInputStream :: FilePath -> IO (exists t. InputStream t => t)
Bear in mind you can't even write IO (forall t. whatever) in Haskell.
> would cause any more problems than
> data Foo = forall t. InputStream t => Foo t
> openInputStream :: FilePath -> IO Foo
> What am I missing?
Simply that undefined is not the same as (Foo undefined).
Quite separately, if InputStream happens to look like this:
class InputStream t where
f1 :: t -> something;
f2 :: t -> something';
f3 :: t -> something'';
where none of the somethings refer to t, you'd be better off with a
f1 :: something;
f2 :: something';
f3 :: something'';
This is a much better way of doing semi-OOP. AFAIK you can't really do
proper OOP-style extensibility in Haskell at all (and "exists" wouldn't
Ashley Yakeley, Seattle WA
More information about the Haskell