make a small example which illustrates the problem and we'll take a look a it. It's not clear to me that one should necessarily expect that the output of an archive class should be identical to it's polymorphic variant. Robert Ramey Maxime van Noppen wrote:

On 08/07/2010 03:54 PM, Maxime van Noppen wrote:

...

I looked at various archive implementations but couldn't find any code that I could relate to this problem.

Little bump. I still get the same behavior with Boost 1.44 so it seems I'm doing something wrong but I really can't find what. :-/ Any idea?

Thanks,

Maxime van Noppen wrote:

On 08/18/2010 04:04 PM, Robert Ramey wrote:

...

make a small example which illustrates the problem and we'll take a look a it.

In the first post I attached such an example.

I went back and looked at your example. You're on the wrong track here. First, classes to be serialized should be not dependent in anyway upon any archive class. So use: class unit { public: unit() : id_(0) { } unit(int id) : id_(id) { } template <typename Archive> void serialize(Archive& ar, const unsigned int version) { ar & BOOST_SERIALIZATION_NVP(id_); } private: int id_; }; note that "polymorphism" is not implemented via virtual functions in the archive class itself so your archive should be dependent on whether or note you're using the polymorphic_archive template. class test_oarchive : public boost::archive::detail::common_oarchive<test_oarchive> { friend class boost::archive::save_access; friend class boost::archive::detail::interface_oarchive<test_oarchive>; public: test_oarchive(std::ostream& out, unsigned int = 0) : out_(out){ } void save_binary(const void* address, std::size_t count) { out_.write(static_cast<const char*>(address), count); } template <typename T> void save(const T& t){ out_ << t; } template <class T> void save_override(const boost::serialization::nvp<T>& nvp, int f) { out_ << "<name=" << nvp.name() << ">"; typedef boost::archive::detail::common_oarchive<test_oarchive> detail_common_oarchive; this->detail_common_oarchive::save_override(nvp.const_value(), f); out_ << "</name=" << nvp.name() << ">\n"; } void save_override(const boost::archive::object_id_type & t, int) { } void save_override(const boost::archive::object_reference_type & t, int) { } void save_override(const boost::archive::version_type & t, int) { } void save_override(const boost::archive::class_id_type & t, int) { } void save_override(const boost::archive::class_id_optional_type & t, int) { } void save_override(const boost::archive::class_id_reference_type & t, int) { } void save_override(const boost::archive::class_name_type & t, int) { } void save_override(const boost::archive::tracking_type & t, int) { } private: std::ostream& out_; }; Test the above to verify that it works according to your taste. Now use a "templated typedef" to generate the polymorphic version: #include <boost/config.hpp> #include <boost/archive/xml_oarchive.hpp> #include <boost/archive/detail/polymorphic_oarchive_route.hpp> typedef boost::archive::detail::polymorphic_oarchive_route< test_oarchive

polymorphic_test_oarchive;

// required by export BOOST_SERIALIZATION_REGISTER_ARCHIVE( test_oarchive ) or something like this. Now you should have test_oarchive and polymorphic_test_oarchive Robert Ramey

...

So now 'unit' only has a template serialize method, but I still

...

int main() { unit u(42);

{ std::cout << "template:" << std::endl; test_oarchive ar(std::cout); ar << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; }

{ std::cout << "polymorphic:" << std::endl; polymorphic_test_oarchive ar(std::cout); ar << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; } }

Which outputs:

...

template: <name=u><name=id_>42</name=id_> </name=u>

polymorphic: 42

Code is attached.

what happens when you try: std::cout << "polymorphic:" << std::endl; polymorphic_oarchive & po = polymorphic_test_oarchive ar(std::cout); po << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; This would actually be using the polymorphic interface as it's intended to be used by routing the serialization through the polymorphic interface. Robert Ramey

Hi, I have a problem with serializing a derived classes through a pointer to a base class. I tried registering with the syntax from the documentation: |ar.template register_type<derived_one>(); | , which works well in case the derived class has a default constructor (just like the example). I just don't know what the syntax is for the case where the derived class does not have a default constructor (if I use the same syntax to register the derived class then I get a compilation error about the fact that there is no. I can provide an example in case this is not clear enough. Is it an absolute requirement for all derived classes to have a default constructor? Thank you and best regards, Dante