Yes. 
I did an example like this to learn how to serialize in the "classical" way. 

My problem is that i can't define a serialize method inside Container, but i would serialize it anyway. How can i serialize the Example strut entirely?

Do you understand my problem?

On Fri, Apr 30, 2010 at 3:49 AM, Robert Ramey <ramey@rrsd.com> wrote:
Marco Meoni - Sbaush wrote:
> Can someone help me?
> Robert, where can I find what you're talking about?
 
I don't know what you're trying to do so I'll take a guess.
 
Does this look like what you want to do?
 
struct Container {
    int m_container_member;
    ... // more members
    template<class Archive>
    void serialize(Archive & ar, const unsigned int version){
        ar & m_container_member;
        ... // more members
    }
};
   
struct Example {
    friend class boost::serialization::access; // not really necessary as all "struct" members are public
    template<class Archive>
    void serialize(Archive & ar, const unsigned int version){
        ar & m_container;
    }
};
 
int main(int argc, char * argv[]){
    const Container c = ...
    std::fostream os("filename");
    boost::serialization::archive ar(os);
    ar & c;
}
 
Does that help any?
 
Robert Ramey
 

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Marco Meoni - Sbaush
www.marcomeoni.net