Copying a C++ class with a member variable of reference type

Question

I\'ve a class which stores a reference to its parent, the reference is passed in the constructor. If I try to copy an instance I get an error \"error C2582: \'operator =\' f

Answer 1

I would make it a boost::shared_ptr. You can be pretty rough with these and they take care of themselves. Whereas using a raw pointer means tha you have to worry about that object being kept alive

Answer 2

I don't recommend this at all

but if you are really gung ho about doing this:

#include <new>

MyClass::MyClass(const MyClass &rhs): parent(rhs.parent)
{
}

MyClass &MyClass::operator=(const MyClass &rhs)
{
    if (this!=&rhs)
    {
        this->~MyClass();
        new (this) MyClass(rhs);
    }

    return *this;
}

Answer 3

Yes, if you need to support assignment, making it a pointer instead of a reference is nearly your only choice.

Answer 4

You need to implement a copy constructor and initialize the reference in that copy constructor, to point to the same reference as the original object.

Answer 5

There is a way to do it and still use a reference, use a reference_wrapper. So

    T& member;

becomes

    std::reference_wrapper<T> member;

Reference wrappers are basically just re-assignable references.

Answer 6

As mentioned by others, using std::reference_wrapper can be used. The helper functions std::ref() and std::cref() can be used, too. Unlike other postings, C++03 introduced reference_wrapper, ref() and cref() in the namespace std::tr1, so you have options if you're not using C++11 or beyond.