如何在另一个名称空间中从全局名称空间定义一个朋友类? [英] How do I define a friend class from the global namespace in another namespace?

问题描述

在先前的问答中(

In a previous Q&A (How do I define friends in global namespace within another C++ namespace?), the solution was given for making a friend function definition within a namespace that refers to a function in the global namespace.

我对类有相同的问题.

class CBaseSD;

namespace cb {
class CBase
{
    friend class ::CBaseSD; // <-- this does not work!?
private:
    int m_type;
public:
    CBase(int t) : m_type(t) {};
};
}; // namespace cb

class CBaseSD
{
private:
    cb::CBase*  m_base;
public:
    CBaseSD(cb::CBase* base) : m_base(base) {};
    int* getTypePtr()
    { return &(m_base->m_type); };
};

如果我将CBaseSD放在命名空间中，它将起作用.例如.， 朋友类SD :: CBaseSD; 但我没有找到适用于全局名称空间的咒语.

If I put CBaseSD into a namespace, it works; e.g., friend class SD::CBaseSD; but I have not found an incantation that works for the global namespace.

我正在使用g ++ 4.1.2进行编译.

I am compiling with g++ 4.1.2.

推荐答案

如问题下方的某些注释所述，问题中的代码似乎可以与我一起使用(Linux-Ubuntu-16.04，gcc版本5.4.0) )，前提是转发声明这样的朋友班.

As stated in some of the comments below the question, the code in the question appears to work with me (Linux-Ubuntu-16.04, gcc version 5.4.0), provided that the friend class was forward-declared.

在寻求答案时，我遇到了

In pursuit of an answer, I came across this post that both explains proper technique for making friend class of a global namespace and answers why it needs to be declared the way it does. It is a nice thread because it references the standard.

如前所述，全局命名空间的类必须先转发声明，然后才能用作命名空间中的类的朋友类.

As stated earlier, the class of a global namespace must be forward declared before it can be used as a friend class to a class within a namespace.

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