使用boost::serialization序列化为XML

这是个新手问题。我正试图将一些对象序列化为XML,但得到的XML包含一个我不需要的增强序列化签名、版本信息、类ID等等。有没有一种方法可以在不后期处理XML消息的情况下消除它们?

#include <fstream>
#include <iostream>
#include <boost/archive/xml_iarchive.hpp>
#include <boost/archive/xml_oarchive.hpp>

using namespace std;

class Test {
private:    
    friend class boost::serialization::access;
    template<class Archive> void serialize(Archive & ar,
            const unsigned int version) {
        ar & BOOST_SERIALIZATION_NVP(a);
        ar & BOOST_SERIALIZATION_NVP(b);
        ar & BOOST_SERIALIZATION_NVP(c);
    }

    int a;
    int b;
    float c;
public:
    inline Test(int a, int b, float c) {
        this->a = a;
        this->b = b;
        this->c = c;
    }
};

int main() {
    std::ofstream ofs("filename.xml");

    Test* test = new Test(1, 2, 3.3);

    boost::archive::xml_oarchive oa(ofs);
    oa << BOOST_SERIALIZATION_NVP(test);

    return 0;
}

结果:

  <?xml version="1.0" encoding="UTF-8" standalone="yes" ?> 
  <!DOCTYPE boost_serialization (View Source for full doctype...)> 
  <boost_serialization signature="serialization::archive" version="6">
  <test class_id="0" tracking_level="1" version="0" object_id="_0">
    <a>1</a> 
    <b>2</b> 
    <c>3.3</c> 
  </test>
  </boost_serialization>

不过,我将把这些消息序列化为字符串,并将它们发送到希望消息看起来像这样的系统。

  <test>
    <a>1</a>
    <b>2</b> 
    <c>3.3</c> 
  </test>

有没有一种方法可以在没有签名的情况下序列化XML?