派生类中受保护的成员函数地址不可访问

问题描述

#include <iostream>

class A {
protected:
    void foo()
    {}
};

class B : public A {
public:
    void bar()
    {
       std::cout << (&A::foo) << std::endl;
    }
};

int main()
{
    B b;
    b.bar();
}

在这里，我试图获取基类的受保护成员函数的地址.我收到此错误.

Here I am trying to get address of protected member function of base class. I am getting this error.

main.cpp: In member function ‘void B::bar()’:
main.cpp:5: error: ‘void A::foo()’ is protected
main.cpp:13: error: within this context
make: *** [all] Error 1

将foo更改为公共作品.也可以打印&B::foo.您能解释一下为什么我们无法获得基类的受保护成员函数的地址吗?

Changing foo to public works. Also printing &B::foo works. Can you please explain why we can't get address of protected member function of base class?

推荐答案

B访问A的受保护成员.在您的示例中，您尝试通过A访问foo，在这种情况下，B是否从A派生无关紧要.

B is allowed to access protected members of A as long as the access is performed through an object of type B. In your example you're trying to access foo through A, and in that context it is irrelevant whether B derives from A or not.

从N3337开始，§11.4/1 [受保护的类]

From N3337, §11.4/1 [class.protected]

 class B {
 protected:
   int i;
   static int j;
 };
 class D1 : public B {
 };
 class D2 : public B {
   friend void fr(B*,D1*,D2*);
   void mem(B*,D1*);
 };
 // ...
 void D2::mem(B* pb, D1* p1) {
   // ...
   int B::* pmi_B = &B::i; // ill-formed
   int B::* pmi_B2 = &D2::i; // OK
   // ...
 }
 // ...

-结束示例]

您的示例与D2::mem中的代码非常相似，这表明尝试通过B而不是D2形成指向受保护成员的指针是错误的.

Your example is very similar to the code in D2::mem, which shows that trying to form a pointer to a protected member through B instead of D2 is ill-formed.

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