Make your own archive derived from xml_?archive.Robert Ramey"Paul Heil" <paul.heil@gmail.com> wrote in message news:2fc4ff370902051118v6b811275rdad4dd46c04d40c9@mail.gmail.com...I'd like to use the boost::serialization library to generate XML that must be read by a third party application. Unfortunately, this application does not like the extra stuff that the serialization library generates.
The code below generates XML that looks like this:
<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>
<!DOCTYPE boost_serialization>
<boost_serialization signature="serialization::archive" version="5">
<MYTAG class_id="0" tracking_level="0" version="0"></MYTAG>
</boost_serialization>
What I need is XML that looks more like this:
<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>
<MYTAG></MYTAG>
What can I do to eliminate the extra stuff?
Thanks,
PaulH
class MyClass
{
private:
friend class boost::serialization::access;
template<class Archive>
void serialize(Archive &ar, const unsigned int version){};
public:
MyClass(){};
friend std::ostream & operator<<(std::ostream &os, const MyClass &gp);
virtual ~MyClass(){};
};
int main( int argc, _TCHAR* argv[] )
{
MyClass mc;
std::ofstream ofs("MyFile.xml");
boost::archive::xml_oarchive oa(ofs);
oa << boost::serialization::make_nvp( "MYTAG", mc );
return 0;
}
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