Hi all, Does anybody know why this is the case? Is there a way to make a class "polymorphic" without adding a dummy virtual function? Is this some "silly" optimization done by C++ compilers? Not generating vtables? It took me quite some time to figure this one out. It is mentioned in the documentation (2 lines, easily missed). http://www.boost.org/doc/libs/1_37_0/libs/serialization/doc/serialization.ht... Maybe add a big huge WARNING? Also please mention this fact in the examples! (like in demo.cpp?) regards M. Seiler #include #include #include #include #include #include #include #include class A { public: A() : test(33) { } int test; virtual void dummy(){} // if you comment this line it won't work }; class B : public A { public: B() : test_b(52) { } int test_b; virtual void foo(){}; }; namespace boost { namespace serialization { template<class Archive> void serialize(Archive & ar, A & o, const unsigned int version) { ar & make_nvp("test", o.test); } template<class Archive> void serialize(Archive & ar, B & o, const unsigned int version) { ar & boost::serialization::make_nvp("A", boost::serialization::base_object<A>(o)); ar & make_nvp("test_b", o.test_b); } } } BOOST_CLASS_EXPORT(B); // Should register B for all archives such that it can be serialized of the basepointers void main() { A* a = new B(); boost::archive::xml_oarchive archive(std::cout); archive.template register_type<B>(); // Should register B for this archive archive & boost::serialization::make_nvp ("a", a); // If you check the output, you see that `{B}.test_b' was not serialized int i; std::cin >> i; }