c++ - 指针的Boost序列化将保存指针的十六进制值，而不是对象的内容

“每当我们在指针(或引用)上调用序列化时，都会在必要时触发其指向(或引用)的对象的序列化”-codeproject.com上的A practical guide to C++ serialization本文提供了一个很好的解释，以演示如何序列化指针序列化指针所指向的数据，因此我编写了一个代码来尝试:

#include <fstream>
#include <iostream>

#include <boost/archive/text_oarchive.hpp>
#include <boost/archive/text_iarchive.hpp>

class datta {
public:
    int integers;
    float decimals;

    datta(){}
    ~datta(){}

    datta(int a, float b)   {
        integers=a;
        decimals=b;
    }

    void disp_datta()   {

        std::cout<<"\n int: "<<integers<<" float" <<decimals<<std::endl;

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

int main()  {

    datta first(20,12.56);
    datta get;
    datta* point_first;
    datta* point_get;

    point_first=&first;

    first.disp_datta();
    std::cout<<"\n ptr to first :"<<point_first;

//Serialize

    std::ofstream abc("file.txt");
    {
        boost::archive::text_oarchive def(abc);
        abc << point_first;
    }

return 0;
}

运行此代码后，我打开file.txt并找到一个十六进制指针地址，而不是该地址所指向的值，我也编写了反序列化代码:

std::ifstream zxc("file.txt");
{
    boost::archive::text_iarchive ngh(zxc);
    ngh >> point_get;
}

//Dereference the ptr and

get = *point_get;

get.disp_datta();
std::cout<<"\n ptr to first :"<<point_get;

我在这里有细分错误!谁能告诉我如何使它工作？非常感谢!

最佳答案

您将对象写入流中，而不是存档中:)

    boost::archive::text_oarchive def(abc);
    abc << point_first;

尝试

    def << point_first;

代替。结果:

22 serialization::archive 10 0 1 0
0 20 12.56

看到它 Live (with deserialization) On Coliru