c++ - Boost XML 基于 class_id 的反序列化

Question

我有一个 XML 序列化反射类型，我可以像这样得到它的序列化:

template <typename Archive>
std::function<void(Archive&,unsigned)> get_serialization_for_type(std::string name);

这些类型都有一个使用 boost 序列化库设置的 GUID，所以它们的 class_id XML 中的属性与有效名称匹配。如何反序列化这些类型？有没有办法可以获取存档正在读取的当前节点的属性？也欢迎就不同方法提出建议，但我无法更改 XML 的格式。

示例 XML:

<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>
<!DOCTYPE boost_serialization>
<boost_serialization signature="serialization::archive" version="17">
   ...
   <container class_id="23" tracking_level="0" version="0">
      <count>2</count>
      <typeA class_id="type_a" tracking_level="0" version="0">
         ...
      </typeA>
      <typeB class_id="type_b" tracking_level="0" version="0">
         ...
      </typeB>
   </container>
   ...
</boost_serialization>

Answer 1

That looks like a regular Boost Serialization XML archive. Why don't you read it in exactly the same way as you wrote it?

我的意思基本上是这样的:这段代码序列化和反序列化具有不同类型和属性的多态元素容器。

它使用“相同”的代码进行序列化和反序列化。

如果 XML 确实是由 Boost Serialization 生成的，原则上您应该能够利用此模式。

Live On Coliru

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

#include <boost/serialization/vector.hpp>
#include <boost/serialization/unique_ptr.hpp>
#include <boost/serialization/assume_abstract.hpp>
#include <boost/serialization/export.hpp>

#include <iostream>
#include <sstream>
#include <boost/core/demangle.hpp>

namespace MyLib {
    struct Base {
        using BaseContainer = std::vector<std::unique_ptr<Base> >;
        virtual ~Base() = default;
        int a, b, c;
    };

    struct A : Base {
        std::string d, e, f;
    };

    struct B : Base {
        float h, i, j;
    };

    template <typename Ar> void serialize(Ar& ar, Base& base, unsigned) {
        ar & BOOST_SERIALIZATION_NVP(base.a)
           & BOOST_SERIALIZATION_NVP(base.b)
           & BOOST_SERIALIZATION_NVP(base.c)
           ;
    }

    template <typename Ar> void serialize(Ar& ar, A& a, unsigned) {
        ar & boost::serialization::make_nvp("Base", boost::serialization::base_object<Base>(a))
           & BOOST_SERIALIZATION_NVP(a.d)
           & BOOST_SERIALIZATION_NVP(a.e)
           & BOOST_SERIALIZATION_NVP(a.f)
           ;
    }

    template <typename Ar> void serialize(Ar& ar, B& b, unsigned) {
        ar & boost::serialization::make_nvp("Base", boost::serialization::base_object<Base>(b))
           & BOOST_SERIALIZATION_NVP(b.h)
           & BOOST_SERIALIZATION_NVP(b.i)
           & BOOST_SERIALIZATION_NVP(b.j)
           ;
    }

    using BaseContainer = std::vector<std::unique_ptr<Base> >;
}

//BOOST_SERIALIZATION_ASSUME_ABSTRACT(MyLib::Base)
BOOST_CLASS_EXPORT(MyLib::Base)
BOOST_CLASS_EXPORT_GUID(MyLib::A, "type_A")
BOOST_CLASS_EXPORT_GUID(MyLib::B, "type_B")

int main() {
    std::stringstream xml;

    {
        MyLib::BaseContainer container;
        container.emplace_back(std::make_unique<MyLib::A>());
        container.emplace_back(std::make_unique<MyLib::B>());
        container.emplace_back(std::make_unique<MyLib::B>());
        container.emplace_back(std::make_unique<MyLib::A>());


        boost::archive::xml_oarchive oa(xml);

        oa << BOOST_SERIALIZATION_NVP(container);
    }

    //std::cout << xml.str();

    {
        boost::archive::xml_iarchive ia(xml);

        MyLib::BaseContainer container;
        ia >> BOOST_SERIALIZATION_NVP(container);

        for (auto& el : container) {
            std::cout << "Element of type " << boost::core::demangle(typeid(*el).name()) << "\n";
        }
    }
}

打印:

Element of type MyLib::A
Element of type MyLib::B
Element of type MyLib::B
Element of type MyLib::A