我正在尝试使用指向基类的指针通过 boost 序列化来使派生类可序列化。
Base::序列化调用。 Derived::serialization 未调用。
我做错了什么?
#include <sstream>
#include <boost/archive/binary_oarchive.hpp>
#include <boost/serialization/serialization.hpp>
#include <boost/serialization/assume_abstract.hpp>
struct Base
{
int x;
Base() { x = 0; }
virtual ~Base() {}
template<class Archive>
void serialize(Archive &ar, const unsigned int version)
{
ar & x;
}
};
BOOST_SERIALIZATION_ASSUME_ABSTRACT(Base)
struct Derived : Base
{
int y;
Derived() { x = 1; y = 2; }
virtual ~Derived() {}
template<class Archive>
void serialize(Archive & ar, const unsigned int version)
{
ar & boost::serialization::base_object<Base>(*this);
ar & y;
}
};
int main()
{
Derived derived;
Base *basePtr = &derived;
std::string s;
std::stringstream ss(s);
boost::archive::binary_oarchive oa(ss);
oa << *basePtr;
}
最佳答案
首先,您没有通过指针序列化,修复它:
oa << basePtr;
其次,您需要注册派生类型:
oa.register_type<Derived>();
或者,在序列化函数中:
ar.template register_type<Derived>();
您可能还想查看注册类信息以进行序列化:http://www.boost.org/doc/libs/1_57_0/libs/serialization/doc/special.html#export
更新评论/编辑的问题:
您的问题代码序列化一个 Base*
并反序列化一个 Derived
。类型不相关,这永远行不通。它们必须相同。
此外,注册读取输入存档的派生类型。
#include <sstream>
#include <boost/archive/binary_oarchive.hpp>
#include <boost/archive/binary_iarchive.hpp>
#include <boost/serialization/serialization.hpp>
#include <boost/serialization/assume_abstract.hpp>
struct Derived;
struct Base
{
int x;
Base() { x = 0; }
virtual ~Base() {}
template<class Archive>
void serialize(Archive &ar, const unsigned int version)
{
ar.template register_type<Derived>();
ar & x;
}
};
BOOST_SERIALIZATION_ASSUME_ABSTRACT(Base)
struct Derived : Base
{
int y;
Derived() { x = 1; y = 2; }
virtual ~Derived() {}
template<class Archive>
void serialize(Archive & ar, const unsigned int version)
{
ar & boost::serialization::base_object<Base>(*this);
ar & y;
}
};
int main()
{
std::stringstream ss;
{
Derived derived;
Base *basePtr = &derived;
boost::archive::binary_oarchive oa(ss);
oa.register_type<Derived>();
oa << basePtr;
ss.flush();
}
{
boost::archive::binary_iarchive ia(ss);
ia.register_type<Derived>();
Base *basePtr = nullptr;
ia >> basePtr;
std::cout << "basePtr->x = " << basePtr->x << "\n";
if (Derived* derivedPtr = dynamic_cast<Derived*>(basePtr))
std::cout << "derivedPtr->y = " << derivedPtr->y << "\n";
}
}
关于c++ - 如何通过指向基类的指针序列化派生类? Derived::serialize 未调用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26964815/