[Haskell-cafe] Existential quantification problem
jonathanccast at fastmail.fm
Sun Jul 13 19:50:50 EDT 2008
On Sun, 2008-07-13 at 18:28 -0500, Derek Elkins wrote:
> On Thu, 2008-07-10 at 10:59 -0700, Jonathan Cast wrote:
> > On Thu, 2008-07-10 at 14:53 -0300, Marco Túlio Gontijo e Silva wrote:
> > > Hello,
> > >
> > > how do I unbox a existential quantificated data type?
> > You can't. You have to use case analysis:
> > case foo of
> > L l -> <whatever you wanted to do>
> > where none of the information your case analysis discovers about the
> > actual type of l can be made available outside of the scope of the case
> > expression. (It can't `escape'). This is required for decidable static
> > typing, IIRC.
> It's not an extraneous requirement; it is part of the definition of
> existential types.
I know that. I didn't know implementing existential types was an end in itself, though.
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