复制赋值运算符应该通过const引用还是按值传递？

Question

Prior to C++11, it has always been the case that copy assignment operator should always pass by const reference, like so:

template <typename T>
ArrayStack<T>& operator= (const ArrayStack& other);

However, with the introduction of move assignment operators and constructors, it seems that some people are advocating using pass by value for copy assignment instead. A move assignment operator also needs to be added:

template <typename T>
ArrayStack<T>& operator= (ArrayStack other);
ArrayStack<T>& operator= (ArrayStack&& other);

The above 2 operator implementation looks like this:

template <typename T>
ArrayStack<T>& ArrayStack<T>::operator =(ArrayStack other)
{
    ArrayStack tmp(other);
    swap(*this, tmp);
    return *this;
}

template <typename T>
ArrayStack<T>& ArrayStack<T>::operator =(ArrayStack&& other)
{
    swap(*this, other);
    return *this;
}

Is it a good idea to use pass by value when creating copy assignment operator for C++11 onwards? Under what circumstances should I do so?

Answer 1

Prior to C++11, it has always been the case that copy assignment operator should always pass by const reference

That is not true. The best approach has always been to use the copy-and-swap idiom , and that's what you're seeing here (although the implementation in the body is sub-optimal).

If anything, this is less useful in C++11 now that you have a move assignment operator too.