> > mapBox :: forall a b. (a -> b) -> Box -> Box > > -- :: forall a b. (a -> b) -> (exists a.a) -> (exists a.a) > > mapBox f (B x) = B (f x) > > > > However, at first sight |f| is polymorphic so it could be applied to > > any value, included the value hidden in |Box|. > > f is not polymorphic here; mapBox is.
I see, it's a case of not paying proper attention. I presume this will reflect in the imposibility of introducing the forall when trying to "prove" the type. > Yes, but that is only because your Box type is trivial. It can contain > any value, so you can never extract any information from it. Indeed, I was just trying to play with unconstrained existentials. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
