Question

如何在struct中序列化std :: allocator类型以供参考，请参阅以下代码作为示例：

// SimpleData.hpp

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

namespace A
{

    struct FullData 
    {
        FullData();
        FullData(const parameter_strings & parms);

        virtual ~FullData();

        std::string country;
        int pincode;

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

         typedef std::list<FullData> FullDataInfo; // what should be done to serialize this data 
    };

}

// SimpleData.cpp
#include "SimpleData.hpp" 
#include <boost/serialization/export.hpp>

namespace A
{
    FullData::FullData()
    {

    }

    FullData::~FullData()
    {

    }

    FullData::FullData(const parameter_strings & parms): detail(parms)
    {

        // impelmentation
    }

}

BOOST_CLASS_EXPORT_IMPLEMENT(A::FullData)
BOOST_CLASS_IMPLEMENTATION(A::FullData    , boost::serialization::object_serializable)
BOOST_CLASS_TRACKING(A::FullData          , boost::serialization::track_never)

我不知道如何序列化 typedef FullData :: inst_list FullDataInfo;

谢谢，

Answer 1

您只需编写代码：

int main()
{
    boost::archive::text_oarchive oa(std::cout);

    A::FullData data1({ { "key1", "value1" }, { "key2", "value2" } });
    data1.pincode = 1234;
    data1.country = "ES";
    A::FullData data2({ { "key3", "value3" }, { "key4", "value4" } });
    data2.pincode = 4321;
    data2.country = "UK";

    A::FullData::FullDataInfo list { data1, data2 };

    oa << list;
}

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输出：

22 serialization::archive 10 0 0 2 0 2 ES 1234 2 UK 4321

完整的代码清单

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

namespace A
{
    typedef std::map<std::string, std::string> parameter_strings;
    struct FullData 
    {
        FullData() : pincode(0) {}
        FullData(const parameter_strings & parms) : pincode(0), detail(parms) {}

        virtual ~FullData() {}

        std::string country;
        int pincode;

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

         parameter_strings detail;

      public:
         typedef std::list<FullData> FullDataInfo; // what should be done to serialize this data 
    };

}

// SimpleData.cpp
#include <boost/serialization/export.hpp>

BOOST_CLASS_EXPORT_IMPLEMENT(A::FullData)
BOOST_CLASS_IMPLEMENTATION(A::FullData    , boost::serialization::object_serializable)
BOOST_CLASS_TRACKING(A::FullData          , boost::serialization::track_never)

int main()
{
    boost::archive::text_oarchive oa(std::cout);

    A::FullData data1({ { "key1", "value1" }, { "key2", "value2" } });
    data1.pincode = 1234;
    data1.country = "ES";
    A::FullData data2({ { "key3", "value3" }, { "key4", "value4" } });
    data2.pincode = 4321;
    data2.country = "UK";

    A::FullData::FullDataInfo list { data1, data2 };

    oa << list;
}

提升std :: list类型变量的序列化不起作用

1 个答案: