Can I add default copies to the constructor?

Question

Can I add default copies to the constructor?

Eg. For this class:

class A
{
    public:
        int a;
        int* b;
};

I want to just write

A::A(const A& rvalue):
    a(rvalue.a),
    b(new int(*(rvalue.b)))
{}

without part a(rvalue.a).

(Ignore bad / ugly code and possible memory leaks)

Answer 1

What you ask for is impossible. Once you declare your own copy constructor, the compiler will not create a copy constructor for you. This means that you cannot simply add or increase the default copy constructor because it will not exist. This is all or nothing, so to speak.

Answer 2

. , , , :

struct A1  {
  int a1;
  int a2;
  // ....
  int aN;
};

struct A:public A1
{
  int* b;
  A(const A& rhs):  A1(rhs), b(new int(*(rhs.b))) {}
};

Answer 3

, , ++: .

, , :

, ( ..), :

//class A hasa large number date members(but all can take advantage of default 
//copy constructor
struct A{
    A(int i):a(i){}
    int a;
    //much more data memberS can use default copy constructor all in class A
};

//class B is simply wrapper for class A 
//so class B can use the default constructor of A
//while just write copy constructor for a raw pointer in it copy constructor
//I think this is what OP want ?
struct B
{
    B(int i,int j):m_a(i),m_b(new int(j)){}

    B(const B & rval):
    m_a(rval.m_a),
    m_b(new int(*rval.m_b))
    {
    }
    A     m_a;
    int * m_b;
};

int main()
{
    B c(2,3); // a=2, *m_b=3
    B d(c);   //after copy constructor, a=2, *m_b=3
}