c++ - 如果我默认复制构造函数，是否会生成 move 构造函数和 move 赋值？

Question

我对如何最好地定义可复制但不可 move 的类感到有点困惑。在我看来，删除 move 构造函数是个坏主意，因为那样我就无法从临时构造函数构造。另一方面，我不希望编译器为我隐式生成一个可能错误或可能导致意外行为的 move 构造函数。我的问题是是否允许编译器为以下类隐式生成 move 构造函数:

class C {
    public:
    int i_;
    C(const C&) = default;
}

Answer 1

[class.copy]/9 在 CWG 1402 之后:

If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared as defaulted if and only if

• X does not have a user-declared copy constructor,
• X does not have a user-declared copy assignment operator,
• X does not have a user-declared move assignment operator, and
• X does not have a user-declared destructor.