er wrote:
If G derives privately from F, and F::result_type is defined, but I provide, say, G::result<F1(T)>,
boost::result_of<G(T)>::type attempts to access the private F::result_type, instead of G::result<G(T)>::type.
I guess, if I really want to keep F a private base, I could wrap it in a class, W<F>, that does not expose result_type, yet I'm hoping someone might have a better idea.
The only solution I can think of is to (partially) specialize boost::result_of directly, as below ... #include <boost/utility/result_of.hpp> struct B { typedef void result_type; }; struct D : private B { template<class S> struct result { typedef int type; }; }; namespace boost { template<typename T> struct result_of<D(T)> : D::result<D(T)> {}; template<typename T> struct result_of<D const(T)> : D::result<D const(T)> {}; // For completeness, specialize for D volatile and // D const volatile, too. } int main() { typedef boost::result_of<D(int)>::type t; } HTH, -- Eric Niebler BoostPro Computing http://www.boostpro.com