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