c++ - 使用 boost::serialization 序列化为 XML

Question

这是一个新手问题。我正在尝试将一些对象序列化为 XML，但生成的 XML 包含增强序列化签名、版本信息、类 id 等。我不需要。有没有办法在不对 xml 消息进行后处理的情况下摆脱它们？

#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，我会将这些消息序列化为字符串，并将它们发送到期望消息看起来像这样的系统。

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

有没有办法在没有签名的情况下序列化xml？

Answer 1

标志 no_header 消除了标题行

unsigned int flags = boost::archive::no_header;
boost::archive::xml_oarchive oa(ofs, flags);

下面的宏消除了属性

BOOST_CLASS_IMPLEMENTATION(Test, object_serializable)