c++ - 在C++17中移动/复制构造函数/赋值是否为 "deleted"或 "not declared"是否成立？

Question

我在看 this answer关于 What are the rules for automatic generation of move operations? ，我希望现在答案已经确定。

幻灯片显示了哪些构造函数/赋值运算符是“未声明的”、“默认的”或“已删除的”，基于已在类中声明的内容，显示:

取自这些 slides ，红色方 block 表示此行为已弃用。

编译时:

#include <iostream>
struct X
{
  template<typename...T>
  X(T&&...) {
    std::cout << "Yay!\n";
  }

  ~X() {}
};

int main() {
  X x0;
  X x1{x0};
  X x2{std::move(x0)};
}

看起来它们已经“未声明”，因为它编译并且输出是“耶!”三遍(这很好，至少对我而言)。但我想确认我可以依赖这种行为。

编辑

已被Frank指出也就是说，如果还添加了复制构造函数，它仍然会说“耶!”三遍，这是一个有趣的行为。做进一步的测试，如果添加移动构造函数，它只会说“耶!”两次。谁能解释这种行为？

Answer 1

根据 N4659(几乎是 C++17 标准)，它们仍被定义为默认，但该行为(仍然)已弃用。

• 复制构造函数，见[class.copy.ctor]/6 :

If the class definition does not explicitly declare a copy constructor, a non-explicit one is declared implicitly. If the class definition declares a move constructor or move assignment operator, the implicitly declared copy constructor is defined as deleted; otherwise, it is defined as defaulted. The latter case is deprecated if the class has a user-declared copy assignment operator or a user-declared destructor.

• 有关复制分配，请参阅 [class.copy.assign]/2 :

If the class definition does not explicitly declare a copy assignment operator, one is declared implicitly. If the class definition declares a move constructor or move assignment operator, the implicitly declared copy assignment operator is defined as deleted; otherwise, it is defined as defaulted. The latter case is deprecated if the class has a user-declared copy constructor or a user-declared destructor.