Serialize to XML using boost::serialization

Question

Hi,

This is a newbie question. I am trying to serialize some objects to XML, but the resulting XML contains a boost serialization signature, version information, class id, ...etc. that I do not need. Is there a way to get rid of them without post-processing the xml message?

#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;
}

results in:

  <?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>

I'll be serializing these messages to strings, though, and sending them to systems that expect a message to look like this.

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

So is there a way to serialize xml without the signature, ...etc.

Thanks,

Answer 1

That is not what boost::serialization should be used for. If you're looking to generate a specific type of XML, better use an XML generator like Xerces (yes, it says "parser" everywhere, but it'll also write XML).