copy constructor in base and derived class

Question

I do not understand two things:

1-In code 1 when the object d2 is defined,default constructor of Base class and copy constructor of Derived class invoked.but in code 2(derived class does not have copy constructor) just copy constructor of Base class invoked!

Why copy constructor of Base class? Is constructor of Derived class called?

2-What is the difference between

 D d2 = d1 ; 

and

D d2;
d2 = d1;

in this example? What is the reason for this?

code 1

//code 1
#include <iostream> 


using std::cout;
using std::endl;

///////////////////////////
class B
{
protected:
    int x;

public:
    B()
    {
        x = 4;
        cout << "default constructor of Base class " << endl;
    }

    B(int a)
    {
        x = a;
        cout << "parametric constructor of Base class" << endl;
    }

    B(const B &b)
    {
        x = b.x;
        cout << "copy constructor of Base class " << endl;
    }
};
////////////////////////////////////////////////////////////
class D : public B
{
private:
    int y;

public:
    D()
    {
        y = 5;
        cout << "default constructor of Derived class" << endl;
    }

    D(int k)
    {
        y = k;
        cout << "parametric constructor of Derived class" << endl;
    }
    D(D& copy)
    {

        cout << "copy constructor of Derived class" << endl;
    }


};
////////////////////////////////////////////////////////////
int main()
{
    D d1(3);

    D d2 = d1;

    return 0;
}

code 2

    //code 2
    #include <iostream> 


    using std::cout;
    using std::endl;

    ///////////////////////////
    class B
    {
    protected:
        int x;

    public:
        B()
        {
            x = 4;
            cout << "default constructor of Base class " << endl;
        }

        B(int a)
        {
            x = a;
            cout << "parametric constructor of Base class" << endl;
        }

        B(const B &b)
        {
            x = b.x;
            cout << "copy constructor of Base class " << endl;
        }
    };
    ////////////////////////////////////////////////////////////
    class D : public B
    {
    private:
        int y;

    public:
        D()
        {
            y = 5;
            cout << "default constructor of Derived class" << endl;
        }

        D(int k)
        {
            y = k;
            cout << "parametric constructor of Derived class" << endl;
        }
 /*     D(D& copy)
        {

            cout << "copy constructor of Derived class" << endl;
        }
 */     

    };
    ////////////////////////////////////////////////////////////
    int main()
    {
        D d1(3);

        D d2 = d1;

        return 0;
    }

Answer 1

Question 1

In "code 1," you've defined a copy constructor for your derived class. Therefore, one is no longer created for you by the compiler. Since you don't specifically call a base constructor in the field initialization list, in your derived class copy constructor, the first thing it does is call the base class default constructor.

However, in "code 2," since you no longer have a defined copy constructor for your derived class, one is implicitly created for you by the compiler. See this question. The top answer explains that the base copy constructor will be called.

Question 2

From this link:

The copy constructor is called whenever an object is initialized (by direct-initialization or copy-initialization) from another object of the same type (unless overload resolution selects a better match or the call is elided), which includes

• initialization: T a = b; or T a(b);, where b is of type T;

So D d2 = d1; calls the D copy constructor, while D d2; d2 = d1; calls the D default constructor, and then the D assignment operator.