Something like this might be made to work. Be aware that the tag name will vary across compilers. For the serialization system this won't be problem as the tag name really isn't used for anything. But, depending on what you plan to use the tag name for, this might or might not be a problem.
Robert Ramey
"Piyush Kapadia" wrote in message news:435b6d4a.334ce710.5f89.ffffb4c0@mx.gmail.com...
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);
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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();
}
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