为什么 C++ 友元类只需要在其他命名空间中进行前向声明? [英] Why does a C++ friend class need a forward declaration only in other namespaces?

问题描述

假设我有一个类 F 应该是类 G(在全局命名空间中)和 C(在命名空间 <代码>A).

Suppose I have a class F that should be friend to the classes G (in the global namespace) and C (in namespace A).

• 要成为 A::C 的朋友，F 必须前向声明.
• 要成为G的朋友，F的前向声明是不必要的.
• 同样，类 A::BF 可以成为 A::C 的朋友，无需前向声明

• to be friend to A::C, F must be forward declared.
• to be friend to G, no forward declaration of F is necessary.
• likewise, a class A::BF can be friend to A::C without forward declaration

以下代码说明了这一点，并可以使用 GCC 4.5、VC++ 10 以及至少一个其他编译器进行编译.

The following code illustrates this and compiles with GCC 4.5, VC++ 10 and at least with one other compiler.

class G {
    friend class F;
    int g;
};

// without this forward declaration, F can't be friend to A::C
class F;

namespace A {

class C {
    friend class ::F;
    friend class BF;
    int c;
};

class BF {
public:
    BF() { c.c = 2; }
private:
    C c;
};

} // namespace A

class F {
public:
    F() { g.g = 3; c.c = 2; }
private:
    G g;
    A::C c;
};

int main()
{
    F f;
}

在我看来，这似乎不一致.这是有原因的还是只是标准的设计决定?

To me this seems inconsistent. Is there a reason for this or is it just a design decision of the standard?

推荐答案

C++ 标准ISO/IEC 14882:2003(E)

7.3.1.2 命名空间成员定义

第 3 段

每个名字首先在一个命名空间是其中的一个成员命名空间.如果朋友声明在一个非本地类首先声明一个类或函数(这意味着类或函数的名称是不合格的)朋友类或函数是最里面的封闭命名空间.

Every name first declared in a namespace is a member of that namespace. If a friend declaration in a non-local class first declares a class or function (this implies that the name of the class or function is unqualified) the friend class or function is a member of the innermost enclosing namespace.

// Assume f and g have not yet been defined.
void h(int);
template <class T> void f2(T);
namespace A {
   class X {
   friend void f(X);  //  A::f(X) is a friend
      class Y {
         friend void g();  //  A::g is a friend
         friend void h(int);  //  A::h is a friend
         //  ::h not considered
         friend void f2<>(int);  //  ::f2<>(int) is a friend
      };
   };
   //  A::f, A::g and A::h are not visible here
   X x;
   void g() { f(x); }  // definition of A::g
   void f(X) { /* ... */}  // definition of A::f
   void h(int) { /* ... */ }  // definition of A::h
   //  A::f, A::g and A::h are visible here and known to be friends
}

你的 friend class BF; 是在命名空间 A 而不是全局命名空间中的 A::BF 声明.您需要全局先验声明来避免这个新声明.

Your friend class BF; is a declaration of A::BF in namespace A rather than global namespace. You need the global prior declaration to avoid this new declaration.

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