如何使用Boost.serialize序列化派生的模板类？

Question

我想序列化/反序列化以下类:

class Feature{
...
virtual string str()=0;
};

template<typename T>
class GenericFeature : public Feature{
T value;
...
virtual string str();
};

我读了boost.serialize docs,并说你必须注册类.我可以在构造函数中注册它们.但是加载会有问题,因为注册是动态的,而不是静态的(据我所知,你必须在序列化/反序列化之前注册类).

如何保存/加载这些类？

Answer 1

首先告诉提示功能是抽象的,并不总是需要:

BOOST_SERIALIZATION_ASSUME_ABSTRACT(Feature);

序列化方法应该看起来或多或少像这样:

template<class Archive> 
void Feature::serialize(Archive & ar, const unsigned int version)
{
   ar & BOOST_SERIALIZATION_NVP(some_member);
}


template<typename T,class Archive> 
void GenericFeature<T>::serialize(Archive & ar, const unsigned int version)
{
   ar & boost::serialization::base_object<Feature>(*this);  //serialize base class
   ar & BOOST_SERIALIZATION_NVP(some_other_member);
}

现在最棘手的一点是在序列化/反序列化中注册类:

boost::archive::text_iarchive inputArchive(somesstream);

boost::archive::text_oarchive outputArchive(somesstream);

//something to serialize
Feature* one = new GenericFeature<SomeType1>();
Feature* two = new GenericFeature<SomeType2>();
Feature* three = new GenericFeature<SomeType3>();

//register our class, must be all of posible template specyfication    
outputArchive.template register_type< GenericFeature<SomeType1> >();
outputArchive.template register_type< GenericFeature<SomeType2> >();
outputArchive.template register_type< GenericFeature<SomeType3> >();

// now simply serialize ;-]
outputArchive << one << two << three;

// register class in deserialization
// must be the same template specification as in serialize
// and in the same correct order or i'm get it wrong ;-D   
inputArchive.template register_type< GenericFeature<SomeType1> >();
inputArchive.template register_type< GenericFeature<SomeType2> >();
inputArchive.template register_type< GenericFeature<SomeType3> >();

Feature* another_one;
Feature* another_two;
Feature* another_three;

// and deserialize ;-]
inputArchive >> another_one >> another_two >> another_three;

如果你需要在某个地方隐藏显式注册并使其更加自动化,那么有想法制作一个特殊的仿函数模板来注册一个派生类,创建所有可用的并放入一个列表,类Feature的一个静态方法会将它们全部注册.但问题是你需要注册所有版本的存档,现在我不知道多态存档是否能完成这项工作.