Re: [Boost-users] boost::serialization - prevent object creation whenloading a pointer

From: Robert Ramey (ramey_at_[hidden])
Date: 2007-12-11 21:33:24

How about the following:

template<class Archive>
 void save(Archive& ar, const unsigned int version) const {
      ar <<  _m;
       cout << "serialise A, _m = " << (unsigned long)_m << endl;
}
void save(Archive& ar, const unsigned int version) const {
    base * t;
     ar >> t; // deserialize to a temporary
    // modify _m according to t
     cout << "serialise A, _m = " << (unsigned long)_m << endl;
    delete t;
}

Robert Ramey

Darryl Lawson wrote:
> I realise that what I am doing is no doubt an unconventional usage of
> boost::serialization, but I wonder if any one has done a similar thing
> before.
>
> I wish to serialize a derived class through a baseclass pointer, but
> on deserilize I do not want the object to be created (I do not want
> memory allocated) - as the object will have all ready been created.
>
> An example, hopefully showing what I mean, is given at the end. The
> program will output something like,
>
>    construct A, _m = 155640808
>    serialise A, _m = 155640808
>    construct A, _m = 155649400
>    serialise A, _m = 155649808
>
> But essentially I want the program to output something lke,
>
>    construct A, _m = 155640808
>    serialise A, _m = 155640808
>    construct A, _m = 155649400
>    serialise A, _m = 155649400
>
> i.e. the second lot of construct A/serialise A will have the same
> address for _m.
>
> The reason why I need this behaviour, is that only a sub-set of the
> derived class will be seriliazed, with the other parts of the class
> intialised differently depending on the session, and the
> initialisation
> of the class is always done _before_ deserialisation.
>
> Is there anyway to get this behaviour from the boost::serialization
> library?
>
> Many thanks,
> Darryl
>
>
> Example
> -------
>
> #include <fstream>
> #include <boost/archive/text_oarchive.hpp>
> #include <boost/archive/text_iarchive.hpp>
> #include <boost/serialization/export.hpp>
>
> using namespace std;
>
> struct base {
>    virtual ~base() {}
>    template<class Archive>
>    void serialize(Archive& ar, const unsigned int version) {}
> };
>
> struct derived : public base {
>    derived() : base(), _value(0) {}
>
>    template<class Archive>
>    void serialize(Archive& ar, const unsigned int version) {
>        boost::serialization::void_cast_register<derived, base>();
>        ar & _value;
>    }
>
>    int _value;
> };
>
> BOOST_CLASS_EXPORT(derived);
>
> struct A {
>    A() {
>        _m = new derived;
>        cout << "construct A, _m = " << (unsigned long)_m << endl;
>    }
>
>    template<class Archive>
>    void serialize(Archive& ar, const unsigned int version) {
>        // On deserialize this will create a new object, but I do not
> want it to,
>        // rather I would like it to deserialize into the all ready
>        created // object.
>        ar & _m;
>        cout << "serialise A, _m = " << (unsigned long)_m << endl;
>    }
>
>    base* _m;
> };
>
> void save(const A& a, const char* filename) {
>    std::ofstream ofs(filename);
>    boost::archive::text_oarchive oa(ofs);
>    oa << a;
> }
>
> void load(A& a, const char* filename) {
>    std::ifstream ifs(filename);
>    boost::archive::text_iarchive ia(ifs);
>    ia >> a;
> }
>
> int main(int argc, char* argv[]) {
>    A a;
>    save(a, "archive.txt");
>    A new_a;
>    load(new_a, "archive.txt");
>    return 0;
> }