Question

请帮我将派生类反序列化为基类指针。我附上了完整的源代码示例。

request.hpp （没有配对cpp文件）

#ifndef REQUEST_HPP
#define REQUEST_HPP

#include <memory>
#include <string>

#include <boost/archive/xml_oarchive.hpp>
#include <boost/archive/xml_iarchive.hpp>

namespace demo {
namespace common {

        class request {
        public:
            static const int INVALID_ID = -42;

            request() 
                : id_(INVALID_ID), timestamp_(0), source_ip_("unknown") {};

            request(int id, long timestamp, const std::string& source_ip) 
                : id_(id), timestamp_(timestamp), source_ip_(source_ip) {};

            virtual ~request() {};

            int id() const { return id_; }
            long timestamp() const { return timestamp_; }
            std::string source_ip() const { return source_ip_; }



        protected:
            int id_;
            long timestamp_;
            std::string source_ip_;



        private:
            friend class boost::serialization::access;

            template<class Archive>
            void serialize(Archive& ar, const unsigned version) {
                ar & BOOST_SERIALIZATION_NVP(id_);
                ar & BOOST_SERIALIZATION_NVP(timestamp_);
                ar & BOOST_SERIALIZATION_NVP(source_ip_);
            }

        };

        typedef std::shared_ptr<request> request_ptr;

    }
};

#endif

command.hpp （派生类）

#ifndef COMMAND_HPP
#define COMMAND_HPP

#include <memory>
#include <string>

#include <boost/serialization/export.hpp>

#include <demo/common/request.hpp>

namespace demo {
    namespace common {

            class command : public request {
            public:
                command(): name_("untitled") {};
                explicit command(const std::string& name) : name_(name) {};
                virtual ~command() {};

                virtual void execute();

                std::string name() const { return name_; }

            protected:
                std::string name_;

            private:
                friend class boost::serialization::access;

                template<class Archive>
                void serialize(Archive& ar, const unsigned version) {
                    ar & BOOST_SERIALIZATION_BASE_OBJECT_NVP(request);
                    ar & BOOST_SERIALIZATION_NVP(name_);
                }

            };

            typedef std::shared_ptr<command> command_ptr;

        }
};

BOOST_CLASS_EXPORT_KEY(demo::common::command)


#endif

command.cpp

#include "command.hpp"
#include <iostream>

BOOST_CLASS_EXPORT_IMPLEMENT(demo::common::command)

namespace demo {
    namespace common {

            void command::execute() {
                std::cout << "  I am '" + name_ +"' and I am executing..." << std::endl;
            }

    }
};

serializer.hpp

#ifndef SERIALIZER_HPP
#define SERIALIZER_HPP

#include <sstream>
#include <string>

/* classes to serialize */
#include <demo/common/request.hpp>
#include <demo/common/command.hpp>

namespace demo {
    namespace common {

        class serializer {
        public:
            serializer() : {};

            template<typename T>
            std::string serialize(const T& t){  
                std::stringstream stream;
                boost::archive::xml_oarchive archive(stream);
                archive << BOOST_SERIALIZATION_NVP(t);
                std::string serialized = stream.str();

                return serialized;
            }


            template<typename T>
            void deserialize(const std::string& serialized, T& t) {
                std::stringstream stream(serialized);
                boost::archive::xml_iarchive archive(stream);
                archive >> BOOST_SERIALIZATION_NVP(t);
            }
        };

    }
}

#endif

样本使用

#include <iostream>

#include <demo/common/serializer.hpp>
#include <demo/common/command.hpp>


using namespace std;
using namespace demo::common;

int main(){
    serializer serializer_;

    command r("123"); // <-- (1) my desired way of declaring
    //request* r = new command("123"); <-- (2) replacing with this makes all work!
    //command* r = new command("123"); <-- (3) replacing with this crashes the app, like (1)
    std::string s = serializer_.serialize(r);
    std::cout << s << std::endl;
    request* rr = nullptr;
    serializer_.deserialize(s, rr); //this throws an exception

    command* rrr = dynamic_cast<command*>(rr);

    rrr->execute();
}

我以为我做了所有需要完成的事情，在任何类导出之前都包含了档案，所有默认构造函数都初始化成员..

请注意，可序列化类和序列化程序被编译为lib文件。然后，该lib用于两个子项目，这些子项目可以访问标题并链接该lib。他们使用这些类相互通信，通过网络发送序列化对象。

为什么我不能将派生类反序列化为基类指针？ 我正在使用Boost 1.51和VC11。

Answer 1

<强>问题：

我发现的两个主要问题很糟糕，而且没有足够的关于引起我问题的Boost :: serialization的文档记录如下：

堆栈上的对象的序列化/反序列化与堆上的对象混合。例如，如果从堆栈上的对象序列化，则尝试反序列化为指针（例如，调用load_construct_data＆lt;＆gt; ;）可能发生异常。与反向情况相同。
没有正确链接您的导出。例如，如果您创建序列化模板/类并将它们放在.lib中，则似乎导出可能未在/ exposed中正确链接。这适用于链接，然后使用共享对象/ DLL。

<强>解决方案：

对于＃1，我发现最简单的做法是始终对指针进行序列化/反序列化。甚至堆栈中的对象在序列化时也可以使用临时指针来允许此规则。例如：

// serialize
MyObject myobj;
std::ostringstream oss;
boost::archive::text_oarchive oa(oss);
MyObject* myObjPtr = &myObj;
oa << myObjPtr; // this is different than oa << myObj!!
std::string serialized = oss.str();

// deserialize
MyObject* myNewObjPtr;
std::stringstream iss(serialized);
boost::archive::text_iarchive ia(iss);
ia >> myNewObjPtr; // invokes new, don't forget to delete (or use smart ptrs!!!)

对于＃2，只需创建一个包含所有导出的.cpp文件。将此CPP直接链接到您的模块中。换句话说，你将拥有一堆带有BOOST_CLASS_EXPORT_IMPLEMENT（）的.cpp：

BOOST_CLASS_EXPORT_IMPLEMENT(MyObject);
// ...

更完整的示例：
下面是一个更完整的示例，显示了一些使用非侵入式模板的序列化技巧。侵入式成员方法将非常相似：

<强> MyObject.h

// Can be broken into MyObject.h, MyObject.cpp, MyObjectSerialization.h for example as well. 
// This stuff can live in your .lib
#include <boost/serialization/export.hpp>
#include <boost/archive/text_oarchive.hpp>
#include <boost/archive/text_iarchive.hpp>
// assume this class contains GetSomeMember() returning SomeMemberType
class MyObject { /* ... */ };
BOOST_CLASS_EXPORT_KEY(MyObject);
namespace boost { namespace serialization {
  template<class Archive>
      void serialize(Archive& ar, MyObject& myObj, const unsigned int version)
  {
      ar & myObj.m_someMember;
  }

  template<class Archive>
      inline void save_construct_data(Archive& ar, const MyObject* myObj, const unsigned int version)
  {
    ar & boost::serialization::make_nvp("SomeMemberType", static_cast<const SomeMemberType&>(myObj->GetSomeMember()));
  }

  template<class Archive>
  inline void load_construct_data(Archive& ar, MyObject* myObj, const unsigned int version)
  {
      SomeMemberType t;
      ar & boost::serialization::make_nvp("SomeMemberType", t);
      ::new(myObj)MyObject(t);
  }
} } // end boost::serialization ns

<强> MyObjectExports.cpp

// This file must be explicitly linked into your module(s) that use serialization.
// This means your executable or shared module/DLLs
#include <boost/serialization/export.hpp>
#include <boost/archive/text_oarchive.hpp>
#include <boost/archive/text_iarchive.hpp>
#include "MyObject.h"

BOOST_CLASS_EXPORT_IMPLEMENT(MyObject);

Answer 2

您在演示时可能会收到 input_stream_error ，而在使用您的库时， unregistered_class 异常。这是由于boost在你的情况下自动注册类的方式。

尽管使用了BOOST_CLASS_EXPORT *宏，但在序列化派生对象并反序列化到其基础时，自动注册过程似乎会混淆。

但是，您可以在对存档执行任何i / o操作之前明确注册类：

// ...
boost::archive::xml_iarchive archive(stream);
// register the class(es) with the archive
archive.template register_type<command>();
archive >> BOOST_SERIALIZATION_NVP(t);
// ...

序列化时使用相同的注册顺序。这使得导出宏变得多余。

Boost将派生类反序列化为基类指针

2 个答案: