一个类是否可以有一个构造函数，该构造函数具有一个参数：对同一类型的另一个实例的引用，但不是复制构造函数？
所以我有一个代表小部件树的 Widget 类，我想要有一个小部件的构造函数，它引用父小部件。该类还有一个显式删除的复制构造函数。这是否意味着引用父窗口小部件的构造函数仍被视为复制构造函数？这是否意味着如果我将此类的实例存储在 std::vector 或其他容器中，可能会发生意外的事情？
class Widget
{
public:
Widget();
Widget(Widget& parentWidget); // Is this a copy constructor?
~Widget();
Widget(Widget&&) = delete;
Widget& operator=(Widget&&) = delete;
Widget(const Widget&) = delete;
Widget& operator=(const Widget&) = delete;
};
So I have a class Widget that represents a tree of widgets, I want to have a constructor for widget that takes a reference to the parent widget. The class also has an explicitly deleted copy constructor. Does this mean that the constructor taking a reference to the parent widget is still considered a copy constructor? Does this mean unexpected things could happen if I were to store an instance of this class in a std::vector or some other container?
class Widget
{
public:
Widget();
Widget(Widget& parentWidget); // Is this a copy constructor?
~Widget();
Widget(Widget&&) = delete;
Widget& operator=(Widget&&) = delete;
Widget(const Widget&) = delete;
Widget& operator=(const Widget&) = delete;
};
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是的，它有资格作为复制构造函数，即使没有
const。我不确定标准容器是否会对此感到困惑，但你的程序员同事肯定会。
使用参考以外的东西。
Widget(Widget *parentWidget);可以工作。Yes, it qualifies as a copy constructor, even without the
const.I'm not sure if the standard containers will get confused by it, but your fellow programmers definitely will.
Use something other than a reference.
Widget(Widget *parentWidget);would work.