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;
}
