基于Boost XML class_id的反序列化

Question

我有一个XML序列化的反映类型，我可以这样获得序列化：

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

这些类型都设置了带有boost序列化库的GUID，因此它们在XML中的class_id属性与有效名称匹配。我如何反序列化这些类型？有没有办法获取档案正在读取的当前节点的属性？也欢迎提出关于不同方法的建议，但是我无法更改XML的格式。

样本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

这看起来像是常规的Boost Serialization XML存档。您为什么不以与编写时完全相同的方式阅读它？

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

它使用“相同”代码进行序列化和反序列化。

如果XML确实是由Boost序列化生成的，则原则上您应该能够利用这种模式。

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#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