c++ - 采用 const 参数的默认移动构造函数

Question

定义类时，以下是否有效？

T(const T&&) = default;

我正在阅读移动构造函数 here并且它解释了如何仍然可以隐式声明默认值:

A class can have multiple move constructors, e.g. both T::T(const T&&) and T::T(T&&). If some user-defined move constructors are present, the user may still force the generation of the implicitly declared move constructor with the keyword default.

在页面底部，它提到了缺陷报告 CWG 2171:

CWG 2171 C++14
X(const X&&) = default was non-trivial, made trivial.

也许 wiki 条目有一个错误 CWG 2171仅指复制构造函数，而不是移动构造函数？

Answer 1

来自 n4296 草案:

8.4.2.1 显式默认函数:

A function that is explicitly defaulted shall

(1.1) — be a special member function,

(1.2) — have the same declared function type (except for possibly differing ref-qualifiers and except that in the case of a copy constructor or copy assignment operator, the parameter type may be “reference to non-const T”, where T is the name of the member function’s class) as if it had been implicitly declared, and

(1.3) — not have default arguments.

12.8.10 复制和移动类对象:

The implicitly-declared move constructor for class X will have the form X::X(X&&)

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结果是:

T(const T&&) = default;

无效，因为隐式声明的移动构造函数具有以下形式:

T(T&&)