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.
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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