c++ - 派生类中 protected 成员函数地址不可访问

Question

#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();
}

这里我试图获取基类的 protected 成员函数的地址。我收到此错误。

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(public)工程。打印 &B::foo 也可以。能否请您解释一下为什么我们无法获取基类的 protected 成员函数的地址？

Answer 1

B 被允许访问 A 的 protected 成员，只要访问是通过 B 类型的对象执行的。在您的示例中，您试图通过 A 访问 foo，在这种情况下，B 是否派生自 A< 无关紧要 或不。

来自 N3337，§11.4/1 [class.protected]

An additional access check beyond those described earlier in Clause 11 is applied when a non-static data member or non-static member function is a protected member of its naming class (11.2) As described earlier, access to a protected member is granted because the reference occurs in a friend or member of some class C. If the access is to form a pointer to member (5.3.1), the nested-name-specifier shall denote C or a class derived from C. All other accesses involve a (possibly implicit) object expression (5.2.5). In this case, the class of the object expression shall be C or a class derived from C. [Example:

 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
   // ...
 }
 // ...

—end example]

您的示例与 D2::mem 中的代码非常相似，它表明试图通过 B 而不是 形成指向 protected 成员的指针D2 格式错误。