Re: [Haskell-cafe] Existential quantification problem

10 Jul 2008

      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.

jcc
...
...
{-# LANGUAGE ExistentialQuantification #-}
data L a = forall l. L (l a)
unboxL (L l) = l
is giving me, in GHC:
Inferred type is less polymorphic than expected
      Quantified type variable `l' escapes
    When checking an existential match that binds
        l :: l t
    The pattern(s) have type(s): L t
    The body has type: l t
    In the definition of `unboxL': unboxL (L l) = l
Thanks.

Re: [Haskell-cafe] Existential quantification problem

Jonathan Cast