在 C++ 中，是否可以从同级的第二个派生对象访问第一个派生对象的受保护基础数据成员？

Question

我正在编写一个 C++ 程序，其中我有两个来自基类的派生对象，例如 Derived_1 和 Derived_2。

在处理过程中，第一个对象调整基础数据成员（这里只是模拟使用 Derived_1 的默认构造函数）。然后我想做的是从 Derived_1 对象中读取该数据，并在 Derived_2 的构造函数中使用它来初始化 Derived_2 中的相同成员。

// -std=c++14
#include <iostream>

class Base {
 public:
  int data(void) { return data_; }

 protected:
  int data_{0};
};

class Derived_1 : public Base {
 public:
  Derived_1(void) { this->data_ = 42; }
};

class Derived_2 : public Base {
 public:
   // Derived_2(const Derived_1& a1) { this->data_ = a1.data_; }
};

int main() {
  Derived_1 a1;
  std::cout << "Derived_1 data: " << a1.data() << '\n';

  // Derived_2 a2 = a1;
  // std::cout << "Derived_2 data: " << a2.data() << '\n';
}

如果Derived_2中的构造函数没有注释，就会出现这个错误：

In constructor ‘Derived_2::Derived_2(const Derived_1&)’:
error: ‘int Base::data_’ is protected within this context
Derived_2(const Derived_1& a1) { this->data_ = a1.data_; }
                                                  ^~~~~

我在这里查看了一些关于 SO 试图找到解决方案的相关问题，（例如） Access a derived private member function from a base class pointer to a derived object 和 Why can I access a derived private member function via a base class pointer to a derived object? 但如果我看到了答案，我目前很难找到答案。可能只是我的经验不足。

感谢您帮我解决这个问题。

Answer 1

是的，可以通过将Derived_2 设为friend of Derived_1：

class Derived_1 : public Base {
    //This says that Derived_2 is a friend, which means that Derived_2
    //can access every member in Derived_1
    friend class Derived_2;
public:
    Derived_1(void) { this->data_ = 42; }
};

所以现在你可以这样做了：

class Derived_2 : public Base {
public:
     Derived_2(const Derived_1& a1)
     { 
         //Legal, Derived_2 is a friend of Derived_1, so it can access the
         //protected member 'data_'
         this->data_ = a1.data_; 
     }
};

请注意，现在Derived_2 可以修改Derived_1 的每个数据成员，而不仅仅是data_

Answer 2

第一个错误是基类中声明的data_，派生类无法访问使用。要解决此问题，您需要包含这是您的派生类：

using Base::data_;

现在，OOP 建议每个对象都有与之关联的定义明确的方法，并且没有变量可以直接访问。你可以这样做：

class Derived_2 : public Base {
 public:
   Derived_2(const int& a1) { this->data_ = a1; }
};

并像这样使用它：

Derived_2 a2(a1.data());

或者您可以将Derived_2 声明为Derived_1 的朋友，就像@Rakete1111 在他的回答中提到的那样：

class Derived_1 : public Base {
//This says that Derived_2 is a friend, which means that Derived_2
//can access every member in Derived_1
friend class Derived_2;
public:
    Derived_1(void) { this->data_ = 42; }
};

但是，如果您这样做，Derived_1 中的所有变量都可以直接由Derived_2 访问。

EDIT 1（按要求）：类的用法：

版本 1：

class Derived_1 : public Base {
 using Base::data_;
 public:
  Derived_1(void) { this->data_ = 42; }
};

class Derived_2 : public Base {
 using Base::data_; 
 public:
    Derived_2(const Derived_1& a1) { this->data_ = a1.data(); }
};

版本 2：

//Declaration of Derived_2 here for Derived_1 
//to know that a class with such a name exists
class Derived_2 : public Base;

class Derived_1 : public Base {
 using Base::data_;
 friend class Derived_2;
 public:
  Derived_1(void) { this->data_ = 42; }
};

class Derived_2 : public Base {
 using Base::data_; 
 public:
    Derived_2(const Derived_1& a1) { this->data_ = a1.data_; }
};

Answer 3

派生类只能访问它们自己的基类子对象的受保护成员。他们无权访问其他类的基类受保护成员，即使该基类与其自己的类型相同。

可能的解决方案：

• 声明Base::data_ public。
• 或者将Derived_2声明为Derived_1的朋友。