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
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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