删除复制构造函数或复制赋值运算符算作 "user declared" 吗？

Question

根据 this presentation，如果复制构造函数或复制赋值运算符是 "user declared"，则不会生成隐式移动操作。 delete复制构造函数或复制赋值运算符算作"user declared"吗？

struct NoCopy {
    NoCopy(NoCopy&) = delete;
    NoCopy& operator=(const NoCopy&) = delete;
};

是否会为 NoCopy class 生成隐式移动操作？还是删除相关复制操作算作 "user declared" 从而抑制隐式移动生成？

如果可能的话，我更喜欢参考标准相关部分的答案。

Answer 1

根据您演示文稿的幻灯片 14，已删除的复制构造函数是 "user declared"，因此会抑制移动生成。

Answer 2

术语“用户声明”在标准中没有正式定义。它与特殊成员函数上下文中的“隐式声明”相反。 [dcl.fct.def.default]/4 可以更清楚地说明这个事实，但意图在那里：

Explicitly-defaulted functions and implicitly-declared functions are collectively called defaulted functions, and the implementation shall provide implicit definitions for them (12.1 12.4, 12.8), which might mean defining them as deleted. A special member function is user-provided if it is user-declared and not explicitly defaulted or deleted on its first declaration. A user-provided explicitly-defaulted function (i.e., explicitly defaulted after its first declaration) is defined at the point where it is explicitly defaulted; if such a function is implicitly defined as deleted, the program is ill-formed.

NoCopy(NoCopy&) = delete;和NoCopy& operator=(const NoCopy&) = delete;都是特殊成员函数的声明。由于 you 显式声明它们，而不是允许编译器隐式声明它们，因此它们是用户声明的。因此，这些声明将抑制每个 [class.copy]/9:

的移动构造函数和移动赋值运算符的隐式声明

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,

— X does not have a user-declared destructor, and

— the move constructor would not be implicitly defined as deleted.

删除复制构造函数或复制赋值运算符算作 "user declared" 吗？

Does deleting a copy constructor or copy assignment operator count as "user declared"?

c++

c++11

move

language-lawyer

copy-constructor