Jaewoo Kim wrote:
>
> > Many people have said this, but I still haven't found an instance where
> > existential types are *needed*. For example, if you declare a data type:
>
> How about this? Is the following example possibly accepted in any extended
> Haskell system?
>
> class Sequence seq a where
> empty :: seq a
> cons :: a -> seq a -> seq a
> ...
>
> data Stack a = forall seq.(Sequence seq) => MkStack (seq a)
> ...
> push a (MkStack seq) = MkStack (cons a seq)
> ...
>
> I'd like to use Stack a, not one of (seq a) - possible representations for
> Stack a.
> Without existential type, could the level of abstraction like (Stack a) be
> reached at?
I think I understand what you're looking for. Correct me if I'm
mistaken, but you're trying to define a Stack data type that can be
implemented using any arbitrary Sequence type, right? That is, one Stack
value can be using a standard list to sequence items, whereas another
Stack value can be using a queue data structure. In order to do that
without involving existential types, I would do something like the
following:
data Sequence a = MkSeq {
cons :: a -> Sequence a,
... }
class SequenceClass seq a where
mkSequence :: seq a -> Sequence a
empty :: seq a
data Stack a = MkStack (Sequence a)
...
push a (MkStack seq) = MkStack (cons seq a)
...
instance SequenceClass [] a where
mkSequence list = let
cons a = mkSequence (a:list)
...
in MkSeq cons ...
empty = []
- Michael Hobbs
