在C ++中，是否可以从同级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

I'm writing a C++ program where I have two derived objects from a base class, say, Derived_1 & Derived_2.

During processing, the first object adjusts the base data member (just simulating that using the Derived_1's default constructor here). What I'd like to do is then read that data from the Derived_1 object and use it during Derived_2's constructor to initialize that same member in Derived_2.

// -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';
}

If the constructor in Derived_2 is uncommented, this error occurs:

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_; }
                                                  ^~~~~

I've looked through a number of related questions here on SO trying to find a solution, (for example) Access a derived private member function from a base class pointer to a derived object and Why can I access a derived private member function via a base class pointer to a derived object? but I'm currently having real difficulty spotting the answer if I've seen it. Probably just my inexperience.

Thanks for helping clear this up for me.

Answer 1

Yes, it is possible, by making Derived_2 a 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; }
};

So now you can do this:

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_; 
     }
};

Note that now, Derived_2 can modify every data member of Derived_1 , not just data_

Answer 2

The first error is that data_ declared in the base class is not accessible to the derived classes for usage. To fix this, you need to include this is your derived class:

using Base::data_;

Now, what OOP recommends is that every object has well defined methods associated to it, and no variable is directly accessible. You can either do:

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

and use it like this:

Derived_2 a2(a1.data());

Or you may declare Derived_2 as a friend of Derived_1 , just as @Rakete1111 mentioned in his answer:

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; }
};

However, if you do this, all the variable in Derived_1 will be directly accessible to Derived_2 .

EDIT 1(as requested): The usage of the class:

Version 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(); }
};

Version 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

Derived classes have access only to protected members of their own base class sub object. They don't have access to other classes' bases protected members, even if that base is the same type as its own.

Possible solutions:

• Declare Base::data_ public.
• Or declare Derived_2 as a friend of Derived_1 .