2

是否可以在 boost::serialization 库中反序列化具有引用且没有默认构造函数的(多态)对象?

class Example
{
    int& value;

public:

    Example(int _value): value(_value) {}
    virtual ~Example() {}

    friend class boost::serialization::access;
    template<typename Archive>
    void serialize(Archive & ar, const unsigned int file_version)
    {
        ar & value;
    }
};

class Usage
{
    Example* example;

public:

    Usage(): example(new Example(123)) {}
    ~Usage() { delete example; }

    friend class boost::serialization::access;
    template<typename Archive>
    void serialize(Archive & ar, const unsigned int file_version)
    {
        ar & example;
    }
};

...

// serialize and deserialize object with reference and no default constructor
{
    Usage source;

    std::ostringstream oss;
    boost::archive::text_oarchive oa(oss);
    oa & source;

    Usage target;

    std::istringstream iss(oss.str());
    boost::archive::text_iarchive ia(iss);
    ia & target; // does not compile
}
4

1 回答 1

5

至于非默认可构造对象,我建议在 此处查看项目非默认构造函数 。 您的类可以通过编写自己的函数模板和.
load_construct_datasave_construct_data

于 2011-02-08T21:33:55.617 回答