Question

我不知道如何序列化从另一个列表类型结构派生的结构。以下是我要问的示例代码：

// Nodes.hpp

#include <boost/serialization/serialization.hpp>
#include <boost/serialization/access.hpp> 
#include <boost/serialization/optional.hpp> 
#include <boost/serialization/list.hpp> 

struct Node
{
      std::string firstname;
      std::string lastname;

   private:
      friend class boost::serialization::access;
      template<class Archive>
      void serialize(Archive &ar, const unsigned int version)
      {
         ar & firstname;
         ar & lastname;
      }

};

struct Nodes : public std::list<Node>
{
      std::string address;
      std::string country;

   private:
      friend class boost::serialization::access;
      template<class Archive>
      void serialize(Archive &ar, const unsigned int version)
      {
         ar & address;
         ar & country;
      }

};

BOOST_CLASS_EXPORT_KEY(Node);
BOOST_CLASS_EXPORT_KEY(Nodes);

BOOST_CLASS_EXPORT_IMPLEMENT(Nodes);         
BOOST_CLASS_IMPLEMENTATION(Nodes,boost::serialization::object_serializable);
BOOST_CLASS_TRACKING(Nodes,boost::serialization::track_never);

请查看“public std :: list”行，并告诉我序列化是否正确。

Answer 1

我不知道你想问什么。不过我知道

不建议继承标准容器（它们不是为此而设计的）

如果您，您仍然必须指定您希望序列化基类：

struct Nodes : public std::list<Node>
{
    std::string address;
    std::string country;

private:
    friend class boost::serialization::access;
    template<class Archive>
    void serialize(Archive &ar, const unsigned int version)
    {
        ar & boost::serialization::base_object<std::list<Node> >(*this);
        ar & address;
        ar & country;
    }

};

当然，我建议将list 成员改为 base 。不要忘记加入boost/serialization/list.hpp：看到它 Live On Coliru

#include <boost/serialization/serialization.hpp>
#include <boost/archive/text_oarchive.hpp>
#include <boost/serialization/export.hpp>
#include <boost/serialization/access.hpp>
#include <boost/serialization/base_object.hpp> 
#include <boost/serialization/optional.hpp> 
#include <boost/serialization/list.hpp> 
#include <boost/serialization/shared_ptr.hpp>
#include <list>

struct Node
{
    std::string firstname;
    std::string lastname;

    private:
    friend class boost::serialization::access;
    template<class Archive>
        void serialize(Archive &ar, const unsigned int version)
        {
            ar & firstname;
            ar & lastname;
        }

};

struct Nodes : public std::list<Node>
{
    std::string address;
    std::string country;

    private:
    friend class boost::serialization::access;
    template<class Archive>
        void serialize(Archive &ar, const unsigned int version)
        {
            ar & boost::serialization::base_object<std::list<Node> >(*this);
            ar & address;
            ar & country;
        }

};

BOOST_CLASS_EXPORT_KEY(Node);
BOOST_CLASS_EXPORT_KEY(Nodes);

BOOST_CLASS_EXPORT_IMPLEMENT(Nodes);
BOOST_CLASS_IMPLEMENTATION(Nodes    , boost::serialization::object_serializable);
BOOST_CLASS_TRACKING(Nodes          , boost::serialization::track_never);

int main()
{
    boost::archive::text_oarchive oa(std::cout);
        Nodes nodes;
        nodes.address = "Mr.";
        nodes.country = "Puerto Mazarique";
        nodes.insert(nodes.end(), Node{"John", "Doe"});
        nodes.insert(nodes.end(), Node{"Jane", "Greenwalt"});
        nodes.insert(nodes.end(), Node{"Morgan", "Cheese"});
        nodes.insert(nodes.end(), Node{"Walton", "Whiz"});

        oa << nodes;
}

提升从struct list派生的struct的序列化

1 个答案: