Boost Serialize - Serialize data in a custom way

Question

If I'm using Boost Serialization to serialize an Integer:

#include <boost/archive/text_oarchive.hpp> 
#include <iostream> 

int main() 
{ 
  boost::archive::text_oarchive oa(std::cout); 
  int i = 1; 
  oa << i; 
}

The result will be the following:
22 serialization::archive 5 1

Now I'm curious if and how I could change the way, certain data is serialized. The data does not need to be deserialized, so if that is not possible anymore, it's not a hindering reason to not doing that.

Lets say the above code should create the following output:
integer 11
(The word integer is added and the value will be increased by 10. The archive-header will not be integrated.)

Would that be possible and how could one achieve that? Is Boost Serialization able to let a user do that without modifying the codebase of Serialization?

PS:
The example-code above is copied from the Highscore-Tutorial

Answer 1

You can write your own archive, something like this:

#include <cstddef> // std::size_t
#include <string>
#include <typeid>

template <typename T>
std::string printName() {
  // Unmangle for your platform or just specialise for types you care about:
  return typeid(T).name();
}

//////////////////////////////////////////////////////////////
// class trivial_oarchive
class trivial_oarchive {
public:
    //////////////////////////////////////////////////////////
    // public interface used by programs that use the
    // serialization library
    typedef boost::mpl::bool_<true> is_saving; 
    typedef boost::mpl::bool_<false> is_loading;
    template<class T> register_type(){}
    template<class T> trivial_oarchive & operator<<(const T & t){
        return *this;
    }
    template<class T> trivial_oarchive & operator&(const T & t){
        // No idea why you'd want to add 10, but this does it:
        return *this << printName<T>() << " " << (t+10);
    }
    void save_binary(void *address, std::size_t count){};
};

(Adapted from the documentation)