Re: [Boost-users] How to select the name of class definitioninstead ofvariable name in XML archive

So instead of oa << BOOST_SERIALIZATION_NVP(T); can we use following to have type name printed in xml file ? oa << boost::serialization::make_nvp(typeid(T).name(), T); _____ From: boost-users-bounces@lists.boost.org [mailto:boost-users-bounces@lists.boost.org] On Behalf Of Robert Ramey Sent: Saturday, October 22, 2005 8:40 PM To: boost-users@lists.boost.org Subject: Re: [Boost-users] How to select the name of class definitioninstead ofvariable name in XML archive People have asked this before, its really needed, but as far as anyone knows, there is no portable way to do this. Depending on your needs type_id might return a string you can use. Robert Ramey "Piyush Kapadia" wrote in message news:435aa424.08d6201b.180b.68ff@mx.gmail.com... In following code - archive results in tag name as T as T is used in BOOST_SERIALIZATION_NVP(T) statement. Instead I want actual type name for T, instead of just T as Tag name, how to achieve this ??? template<class T> void Serialize(const char * filename) { std::ofstream ofs(filename); assert(ofs.good()); boost::archive::xml_oarchive oa(ofs); oa << BOOST_SERIALIZATION_NVP(T); ofs.close(); } _____ _______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users