Base and Derived class copy constructor help.

I have

Code:

---

class Base {
int a;

public:
Base();
Base(int);
Base(const Base& b);  //copy constructor
};

Base::Base(const Base& b){
a=b.a;}



class Derived public: Base{
int x;

public:
Derived();
Derived(int);
Derived(const Derived& d); //copy constructor
};


Derived::Derived(const Derived& d) :Base(d){  //<--MY QUESTION HERE.
x=d.x;
}

---

how the Base copy constructor will be executed?
d is a reference to Derived class. It is not a reference to a Base class.
Any help will be appreciated.

Code:

---

Derived::Derived(const Derived& d) :Base(d){

---

'd' will be implicitly converted to its base class 'Base'

With public inheritence, an object of derived class IS an object of the base class. That is why, as kmdv said, a reference to derived can be implicitly converted to a reference to base. (The reverse is not true).

Incidentally, if all you're doing in a copy constructor is copying all members by value, you do not need to declare or define the copy constructor. The compiler will supply one for you that does the right thing (including initialising bases). A similar comment applies for the assignment operator.

there are more variables in the two classes.
one of them is a char*.

Thx for clearing it up for me :)

Consider using std::string or smart pointers then.