我尝试为我的非成员 serialize() 函数提供 A 类的 getter，因为从成员访问是私有的.
I have tried providing getters of class A for my non-member serialize() function` since accessing from members is private.
template<typename T>
class A
{
public:
A(const T& id) : m_id(id) {}
T& getRef() { return m_id; } // not giving good results
T getId() { return m_id; } // not giving good results
const T& getRef() const { return m_id; } // not giving good results
private: // I would like to keep it private
T m_id;
}
namespace boost { namespace serialization {
template<class Archive,typename T>
void serialize(Archive &ar, A &a, const unsigned int version)
{
// ar &BOOST_SERIALIZATION_NVP(a.m_id); // I would like to avoid that it works if m_id is public
ar &BOOST_SERIALIZATION_NVP(a.GetRef()); // I want this !
}
}}
// and later I use
std::ofstream ofs("test.xml");
boost::archive::xml_oarchive oa(ofs);
A<int> a(42);
oa << BOOST_SERIALIZATION_NVP(a);
不幸的是，当我尝试使用 getter GetRef() 或GetId().
如果我在公开时直接访问 m_id，则效果很好.
Unfortunately the execution keeps telling me uncaught exception of type boost::archive::xml_archive_exception - Invalid XML tag name when I try to use getters either GetRef()or GetId().
It works well if I access directly to m_id when it is public.
有什么好的方法吗?
你可以使用老派的好朋友:
You can use good old-fashioned friends:
生活在 Coliru
template <typename T>
class A {
public:
A(const T &id) : m_id(id) {}
private:
template <typename Ar, typename U> friend void boost::serialization::serialize(Ar&,A<U>&,const unsigned);
T m_id;
};
namespace boost {
namespace serialization {
template <class Archive, typename T>
void serialize(Archive &ar, A<T> &a, const unsigned int)
{
ar & BOOST_SERIALIZATION_NVP(a.m_id);
}
}
}
您可以使用 getRef() 方法.这个
make_nvp(因为你不能使用 a.getRef() 作为 XML 元素名称make_nvp (because you can't use a.getRef() as an XML element name遗憾的是，引用 getter 以一种可怕的方式破坏了封装.我个人更喜欢首先公开 m_id.
Sadly, having the reference getter break encapsulation in a horrific way. I'd personally prefer to have
m_idpublic in the first place, instead.
生活在 Coliru
template <typename T>
class A {
public:
A(const T &id) : m_id(id) {}
T& getRef() { return m_id; }
T const& getRef() const { return m_id; }
private:
T m_id;
};
namespace boost {
namespace serialization {
template <class Archive, typename T>
void serialize(Archive &ar, A<T> &a, const unsigned int)
{
ar & boost::serialization::make_nvp("m_id", a.getRef());
}
}
}
您可以使用pimpl"风格的结构.您可以在 A<> 中转发声明一个结构体:
You can use a 'pimpl' style struct. You can forward declare a struct inside A<>:
template <typename T>
class A {
public:
struct access;
A(const T &id) : m_id(id) {}
private:
T m_id;
};
这比 getRef() 方法的侵入性要小，后者只是完全破坏了封装.现在，您可以在该类中隐藏私有访问权限:
That's less intrusive than the getRef() approach which simply breaks encapsulation all the way. Now, you can hide the private access inside this class:
namespace boost {
namespace serialization {
template <class Archive, typename T>
void serialize(Archive &ar, A<T> &a, const unsigned int version)
{
A<T>::access::serialize(ar, a, version);
}
}
}
当然，您仍然需要实现它，但这可以在单独的标题中完成，并且根本不会影响类 A<>(或其任何特化):
Of course you still need to implement it, but this can be done in a separate header and doesn't influence class A<> (or any of its specializations) at all:
template <typename T>
struct A<T>::access {
template <class Archive>
static void serialize(Archive &ar, A<T> &a, const unsigned int) {
ar & BOOST_SERIALIZATION_NVP(a.m_id);
}
};
看看它生活在 Coliru 以及p>
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