c++ - using-declaration 无法正常工作

Question

在以下示例中，我试图隐藏 using Employee::showEveryDept来自最后一个 child 类Designer通过在类里面设为私有(private) Elayer -

#include <iostream>

class Employee {
private:
    char name[5] = "abcd";
    void allDept() { std::cout << "Woo"; }

public:
    void tellName() { std::cout << name << "\n"; }
    virtual void showEveryDept()
    {
        std::cout << "Employee can see every dept\n";
        allDept();
    }
};

class ELayer : public Employee {
private:
    using Employee::showEveryDept;

protected:
    ELayer() {}

public:
    using Employee::tellName;
};

class Designer : public ELayer {
private:
    char color = 'r';

public:
    void showOwnDept() { std::cout << "\nDesigner can see own dept\n"; }
};

int main()
{
    Employee* E = new Designer;
    E->showEveryDept(); // should not work

    Designer* D = dynamic_cast<Designer*>(E);
    D->showOwnDept();
}

但它仍在编译，输出是 -

Employee can see every dept
Woo
Designer can see own dept

但我已明确将其设为私有(private)，请参阅 - private: using Employee::showEveryDept;

我在这里做错了什么？

Answer 1

你想错了。

C++ 有 Name Lookup 的概念，这是一个构建良好的概念，我们通常不会想到，但这在任何地方都会发生 name用来。通过这样做:

int main()
{
    Employee* E = new Designer;
    E->showEveryDept(); // should not work

    Designer* D = dynamic_cast<Designer*>(E);
    D->showOwnDept();
}

E->showEveryDept() 行对属于 E 类的成员(在本例中为成员函数)进行非限定名称查找。因为它是 accessible name , 该程序是合法的。

---

我们也知道 Designer 是从 ELayer 派生的，其中 showEveryDept() 被声明为 private在这里做了:

class ELayer : public Employee {
private:
    using Employee::showEveryDept;

protected:
    ELayer() {}

public:
    using Employee::tellName;
};

但您所做的只是明确地 introduce将 Employee 类中的名称 showEveryDept() 转换为 Elayer；引入 private 访问。这意味着，我们不能在类成员/静态函数之外直接访问与 ELayer 关联的名称(或调用该函数)。

ELayer* e = new Designer();
e->showEveryDept();    //Error, the name showEveryDept is private within ELayer

但是，由于 showEveryDept() 在 Elayer 的基类中具有 public 访问权限，因此 Employer;仍然可以使用 qualified name lookup 访问它

ELayer* e = new Designer();
e->Employer::showEveryDept();    //Ok, qualified name lookup, showEveryDept is public

Elayer 中可在 Designer 中访问的名称将由其 access specification 决定.如您所见，名称 showEveryDept() 是私有(private)的，因此 Designer 甚至不能使用这样的名称。

对于您当前的类层次结构和定义，这意味着:

Employee* E = new Designer();
ELayer*   L = new Designer();
Designer* D = new Designer();

-- 对于 unqualified lookup :

• 这行得通

  E->showEveryDept();                // works!
  

• 这失败了:

  L->showEveryDept();                // fails; its private in Elayer
  

• 这也失败了:

  D->showEveryDept();                // fails; its inaccessible in Designer
  

-- 对于 qualified lookup ，在这种情况下，请求从其基类调用此类函数:

• 这失败了:

  D->Elayer::showEveryDept();        // fails! its private in Elayer
  

• 这行得通:

  D->Employee::showEveryDept();      // works! its accessible in Employee
  

• 这也有效:

  L->Employee::showEveryDept();      // works! its accessible in Employee
  

---

注意:将任何成员函数的名称，B::func(例如)从基类 B 引入派生类 D ，不会覆盖 B::func，它只是让 B::func 在派生类的上下文中对重载决议可见。查看 this question 的答案和 this