Elizabeta wrote:
Hi,
Is it ok to serialize/deserialize in the object I get when dereference a pointer ?
void f() {
B* bObj=new B();
/// save { std::ofstream ofs("C:\\hh.txt"); boost::archive::xml_oarchive oa(ofs); oa & make_nvp("bObj",*bObj); } //// Load { std::ifstream ifs("C:\\hh.txt");
boost::archive::xml_iarchive ia(ifs); ia & make_nvp("bObj",*bObj); } }
My second question : Is it ok to serialize/deserialize in *this ?
Thanks
I'm going to rephrase the question for my own convenience: "How is the serialization library intended to be used to do the equivalent of: stream << *x ?" Here is my answer to my own question. Serializing a data object in a way that most people intend requires that the object recreated be "equivalent or substitutable" for the original object. When an object includes a pointer, it becomes clear that this is not so easy. There are a number of issues the crop up - including a) the fact that pointer to class X might really be refering to some derived class (virtual inheritance) b) there might be multiple pointers refering to the same object. c) same problems for references d) an the fact that all this might be going on simultaneously. The usage of ar << *x or ar << *this is an indication that the user is likely to "re-implement" some of the serialization library. This might occur because the simple interface of the library suggests that it doesn't do much and this is not at all true. Sooooo if you find yourself wanting to do this, step back and think about why you want to do this. Look into the library documentation and examples to see how you might get what you want without doing this. If it's still not clear, it's likely that your overlooking something that the library does that is not obvious. If you still feel you have to do this, I've failed in my task. So the short answer to my phrasing of your question is: "How is the serialization library intended to be used to do the equivalent of: stream << *x ?" ar << x This DOES do ar << *x plus everything else that has to be done to make such an operation make any sense at all. Robert Ramey