Copy_and_base, C++ (vc++)

Copy_and_base

//Microsoft (R) C/C++ Optimizing Compiler Version 19.00.23506 for x86

#include <iostream>

struct Base
{
    char i = 0;
    
    //If you comment out private access specifier below, the code won't compile
    //because Deriv constructor and destructor will implicitly call base class
    //constructor and destructor. So Base class must have their constructor
    //and destructor at least protected to grant access from sub-class.
//private:
    Base(){std::cout << "Base default constructor" << std::endl; }
    virtual ~Base() { std::cout << "Base destructor" << std::endl; }
    
    //whilst, Copy constructor and copy assignment does not needs to be accessed
    //from Derived clases if they were overriden in sub-class.
private:
    Base(const Base&);
    Base& operator=(const Base&);
};

struct Deriv : Base
{
    Deriv(){ std::cout << "Deriv default constructor" << std::endl; }
    ~Deriv() { std::cout << "Deriv destructor" << std::endl; }
    
    //comment out the following copy functions, will generate default implementations
    //that calls base class copy functions. In case of private copy functions in Base
    //class, it will be doomed to fail.
    Deriv(const Deriv&){ std::cout << "Deriv Copy constructor" << std::endl; }
    Deriv& operator=(const Deriv&){ std::cout <<"Deriv Copy assignment" << std::endl; return *this; }

};

int main()
{
   std::cout << "----------" << std::endl;
   std::cout << "sub-class constructor implicitly calls base constructor" << std::endl;
   std::cout << "sub-class destructor implicitly calls base destructor" << std::endl;
   std::cout << "sub-class copy constructor(assignment) also implicitly calls base copy constructor(assignment)" << std::endl;
   std::cout << "    if sub-class did not define its own overridden version of copy constructor(assignment)" << std::endl;
   std::cout << "----------" << std::endl;
   Deriv d;
   std::cout << "----------" << std::endl;
   Deriv d1 = d;
   std::cout << "----------" << std::endl;
   Deriv d2{d1};
   std::cout << "----------" << std::endl;
   d2 = d;
   std::cout << "----------" << std::endl;
}


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