Question

我正在研究一个客户端 - 服务器应用程序，它使用boost :: serialization库来满足它的序列化需求。

我需要序列化和反序列化似乎不起作用的多态对象。文档确实说支持它，但没有相关的例子说明我在这里要做的事情。所以，我不太确定。我的问题是可以使用boost序列化/反序列化多态对象吗？如果是的话，我在这里做错了什么？

谢谢！

代码：

using namespace std;  

class base {  
  public:
    int data1;  

    friend class boost::serialization::access;  

    void serialize(boost::archive::polymorphic_iarchive & ar, 
                   const unsigned int file_version) {  
        ar & data1;  
    }  

    void serialize(boost::archive::polymorphic_oarchive & ar, 
                   const unsigned int file_version){  
        ar & data1;  
    }  

  public:  
    base() {};  
    base(int _d) : data1(_d) {}  
    virtual void foo() const {std::cout << "base" << std::endl;}  
};  

class derived : public base {  
  public:  
    int data2;  

    friend class boost::serialization::access;  

    void serialize(boost::archive::polymorphic_iarchive & ar, 
                   const unsigned int file_version) {  
        ar & boost::serialization::base_object<base>(*this) & data2;  
    }  

    void serialize(boost::archive::polymorphic_oarchive & ar, 
                   const unsigned int file_version){  
        ar & boost::serialization::base_object<base>(*this) & data2;  
    }  

  public:  
    derived() {};  
    derived(int _b, int _d) : base(_b), data2(_d) {}  
    virtual void foo() const {std::cout << "derived" << std::endl;}  
};  

int main(int argc, char *argv[]) {  
    // client  
    const base *b1 = new derived(1, 2);  

    std::ostringstream oss;  
    boost::archive::polymorphic_text_oarchive oa(oss);  
    oa << *b1;  

    // server  
    base *b2 = new derived(3, 4);  

    std::istringstream iss(oss.str());  
    boost::archive::polymorphic_text_iarchive ia(iss);  
    ia >> *b2;  

    // prints 1, ok  
    cout << b2->data1 << endl;  

    // prints 4, why wasn't the derived class data written?
    cout << (dynamic_cast<derived*>(b2))->data2 << endl;  

    return 0;  
}

Answer 1

找到了解决方案。我必须使用语句

导出派生类

BOOST_CLASS_EXPORT(derived);

发布适用于某些更正的内容。

using namespace std;  

class base {  
  public:  
    int data1;  

    friend class boost::serialization::access;  

    template<typename Archive>  
    void serialize(Archive & ar, const unsigned int file_version) {  
        ar & data1;  
    }  

  public:  
    base() {};  
    base(int _d) : data1(_d) {}  
    virtual void foo() const {std::cout << "base" << std::endl;}  
};  

class derived : public base {  
  public:  
    int data2;  

    friend class boost::serialization::access;  

    template<typename Archive>  
    void serialize(Archive & ar, const unsigned int file_version) {  
        ar & boost::serialization::base_object<base>(*this);  
        ar & data2;  
    }  

  public:  
    derived() {};  
    derived(int _b, int _d) : base(_b), data2(_d) {}  
    virtual void foo() const {std::cout << "derived" << std::endl;}  
};  

BOOST_CLASS_EXPORT(derived);  

int main(int argc, char *argv[]) {  
    // client  
    // Assign to base type  
    std::unique_ptr<const base> b1(new derived(1, 2));  

    std::ostringstream oss;  
    boost::archive::text_oarchive oa(oss);  
    oa & b1.get();   

    // server  
    // Retrieve derived type from base  
    std::unique_ptr<base> b2;

    std::istringstream iss(oss.str());
    boost::archive::text_iarchive ia(iss);
    {
        base *temp; 
        ia & temp;
        b2.reset(temp);
    }
    cout << b2->data1 << endl;  
    cout << (dynamic_cast<derived*>(b2.get()))->data2 << endl;  

    return 0;  
}

Answer 2

只是一些评论......

首先，您可以使用相同的操作来使用模板化版本进行序列化和反序列化：

template<class Archive>
void load(Archive & ar, const unsigned int version)
{
   ...
}

此外，您可以“配置”存档以期望这些类型被识别为指针，而不是宏：

ar.register_type(static_cast<your_static_type_here *>(NULL));

使用多态存档提升序列化

2 个答案: