copy constructor is implicitly deleted because the default definition would be ill-formed

Question

I've got a class A (from a library over which I have no control) with a private copy constructor and a clone method, and a class B derived from A. I would like to implement clone for B as well.

The naive approach

#include <memory>

class A { // I have no control here
  public:
    A(int a) {};

    std::shared_ptr<A>
      clone() const
      {
        return std::shared_ptr<A>(new A(*this));
      }

  private:
    A(const A & a) {};
};

class B: public A {
  public:
    B(int data, int extraData):
      A(data),
      extraData_(extraData)
    {
    }

    std::shared_ptr<B>
    clone() const
    {
      return std::shared_ptr<B>(new B(*this));
    }

  private:
    int extraData_;
};

int main() {
  A a(1);
}

however, fails, since the copy constructor of A is private:

main.cpp: In member function ‘std::shared_ptr<B> B::clone() const’:
main.cpp:27:42: error: use of deleted function ‘B::B(const B&)’
     return std::shared_ptr<B>(new B(*this));
                                      ^
main.cpp:17:7: note: ‘B::B(const B&)’ is implicitly deleted because the default definition would be ill-formed:
 class B: public A {
       ^
main.cpp:14:5: error: ‘A::A(const A&)’ is private
     A(const A & a) {};
     ^
main.cpp:17:7: error: within this context
 class B: public A {

There might a way to make use of A::clone() for B::clone(), but I'm not sure how this would work exactly. Any hints?

Answer 1

I presume it's a typo that your B has no public members at all, and that you're missing a public: before the definition of B::B(int,int).

The author of the class represented by your A apparently wants it to be cloneable but not copy constructible. That would suggest he or she wants all instances to live on the heap. But contrariwise, there's the public constructor A::A(int). Are you sure you are right about that?

It's plausible to suppose that the class can reveal enough information about a given instance to constitute another instance. E.g., putting a little more flesh on A:

class A {
public:
    A(int a) 
    : data_(a){};

    std::shared_ptr<A>
    clone() const
    {
        return std::shared_ptr<A>(new A(*this));
    }

    int data() const {
        return data_;
    }

private:
    A(const A & a) {};
    int data_;
};

And if that is true, then the public constructor would render it merely inconvenient to circumvent the private, undefined copy constructor:

A a0(1);
A a1{a0.data()};     // Inconvenient copy construction

So I'm less than confident that A faithfully represents the problem class. Taking it at face value, however, the question you need to answer is: Can you even inconveniently copy construct an A?

If not then you're stuck. If so, then you can use inconvenient copy construction of A to expressly define a conventional copy constructor for B, which is all you need. E.g.

class B: public A {
public:
    B(B const & other)
    : A(other.data()),extraData_(other.extraData_){}    

    B(int data, int extraData):
    A(data),
    extraData_(extraData)
    {
    }

    std::shared_ptr<B>
    clone() const
    {
        return std::shared_ptr<B>(new B(*this));
    }

    int extradata() const {
        return extraData_;
    }

private:
    int extraData_;
};

#include <iostream>

int main()
{
    B b(1,2);
    std::shared_ptr<B> pb = b.clone();
    std::cout << pb->data() << std::endl;
    std::cout << pb->extradata() << std::endl;
    return 0;
} 

Answer 2

You need to make the copy-constructor of A protected so that the derived class could use it:

protected:
    A(const A & a) { /*...*/ }

Hope that helps.