嵌套 类 中的好友声明需要前向声明
Friend declaration in nested classes requiring forward declaration
我正在尝试为具有私有成员的嵌套 class 编写非侵入式 boost::serialization 例程。不幸的是,我没能说服 g++ 序列化例程是内部 class 的朋友。似乎 g++ 需要序列化例程的前向声明,而这又需要嵌套 class 的前向声明,而这又不能在 C++ 中完成。我错过了什么或者这是不可能的吗?相反,clang++ 不需要前向声明,下面的代码也没有问题。下面的代码说明了这个问题:
#include <boost/archive/text_oarchive.hpp>
class Outer;
//class Outer::Inner; // Not valid C++
namespace boost
{
namespace serialization
{
template <class Archive>
void serialize(Archive &ar, Outer& outer, const unsigned int version);
//template <class Archive>
//void serialize(Archive &ar, Outer::Inner& inner, const unsigned int version); // Cannot be done since forward declaration of nested class not possible.
}
}
class Outer
{
class Inner
{
int member_{42};
template <class Archive>
friend void boost::serialization::serialize(Archive &ar, Outer::Inner &inner, const unsigned int version); // This does not work with gcc since the compiler seems to expect a forward declaration, which cannot be done (see above).
};
Inner inner_;
template <class Archive>
friend void boost::serialization::serialize(Archive &ar, Outer &outer, const unsigned int version);
template <class Archive>
friend void boost::serialization::serialize(Archive &ar, Inner &inner, const unsigned int version);
};
namespace boost
{
namespace serialization
{
template <class Archive>
void serialize(Archive &ar, Outer& outer, const unsigned int version)
{
ar & outer.inner_;
}
template <class Archive>
void serialize(Archive &ar, Outer::Inner& inner, const unsigned int version)
{
ar & inner.member_;
}
}
}
int main()
{
Outer outer;
boost::archive::text_oarchive(std::cout) << outer;
}
与 -std=c++11
和 -lboost_serialization
一起编译。使用 g++ 编译会抱怨 member_
是私有的,即使存在友元声明也是如此。 g++ 拒绝内部 class 中的友元声明是否正确?
When the declarator-id is qualified, the declaration shall refer to
a previously declared member of the class or namespace to which the
qualifier refers (or, in the case of a namespace, of an element of the
inline namespace set of that namespace ([namespace.def])) or to a
specialization thereof; [...].
换句话说,具有限定名称的声明(包括友元声明)必须引用先前声明的内容。所以 GCC 在拒绝代码方面是正确的,但它应该更早地拒绝它,并且诊断相当混乱。 (请注意,如果被加为好友的是以前未声明的普通函数而不是模板,它将当场拒绝。)
此外,要求好友访问首先会破坏非侵入式序列化的要点(允许您在不更改其定义的情况下序列化 class)。
我正在尝试为具有私有成员的嵌套 class 编写非侵入式 boost::serialization 例程。不幸的是,我没能说服 g++ 序列化例程是内部 class 的朋友。似乎 g++ 需要序列化例程的前向声明,而这又需要嵌套 class 的前向声明,而这又不能在 C++ 中完成。我错过了什么或者这是不可能的吗?相反,clang++ 不需要前向声明,下面的代码也没有问题。下面的代码说明了这个问题:
#include <boost/archive/text_oarchive.hpp>
class Outer;
//class Outer::Inner; // Not valid C++
namespace boost
{
namespace serialization
{
template <class Archive>
void serialize(Archive &ar, Outer& outer, const unsigned int version);
//template <class Archive>
//void serialize(Archive &ar, Outer::Inner& inner, const unsigned int version); // Cannot be done since forward declaration of nested class not possible.
}
}
class Outer
{
class Inner
{
int member_{42};
template <class Archive>
friend void boost::serialization::serialize(Archive &ar, Outer::Inner &inner, const unsigned int version); // This does not work with gcc since the compiler seems to expect a forward declaration, which cannot be done (see above).
};
Inner inner_;
template <class Archive>
friend void boost::serialization::serialize(Archive &ar, Outer &outer, const unsigned int version);
template <class Archive>
friend void boost::serialization::serialize(Archive &ar, Inner &inner, const unsigned int version);
};
namespace boost
{
namespace serialization
{
template <class Archive>
void serialize(Archive &ar, Outer& outer, const unsigned int version)
{
ar & outer.inner_;
}
template <class Archive>
void serialize(Archive &ar, Outer::Inner& inner, const unsigned int version)
{
ar & inner.member_;
}
}
}
int main()
{
Outer outer;
boost::archive::text_oarchive(std::cout) << outer;
}
与 -std=c++11
和 -lboost_serialization
一起编译。使用 g++ 编译会抱怨 member_
是私有的,即使存在友元声明也是如此。 g++ 拒绝内部 class 中的友元声明是否正确?
When the declarator-id is qualified, the declaration shall refer to a previously declared member of the class or namespace to which the qualifier refers (or, in the case of a namespace, of an element of the inline namespace set of that namespace ([namespace.def])) or to a specialization thereof; [...].
换句话说,具有限定名称的声明(包括友元声明)必须引用先前声明的内容。所以 GCC 在拒绝代码方面是正确的,但它应该更早地拒绝它,并且诊断相当混乱。 (请注意,如果被加为好友的是以前未声明的普通函数而不是模板,它将当场拒绝。)
此外,要求好友访问首先会破坏非侵入式序列化的要点(允许您在不更改其定义的情况下序列化 class)。