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. jcc _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
