默认的移动构造函数采用const参数 [英] Default move constructor taking a const parameter

问题描述

定义类时，以下内容有效吗?

When defining a class, is the following valid?

T(const T&&) = default;

我在此处中了解了移动构造器，并且它解释了默认构造器仍可以隐式声明:

I was reading about move constructors here and it explains how the default may still be implicitly declared:

一个类可以有多个move构造函数，例如两者T::T(const T&&) 和T::T(T&&).如果存在一些用户定义的move构造函数，则 用户仍然可以强制生成隐式声明的动作 关键字为default的构造函数.

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:

At the bottom of the page it mentions defect report CWG 2171:

CWG 2171 C ++ 14
X(const X&&) = default不平凡，变得平凡.

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

也许Wiki条目只是有一个错误，并且 CWG 2171 仅指复制构造函数，而不是move构造函数?

Maybe the wiki entry just has a mistake and CWG 2171 is only referring to a copy constructor, not a move constructor?

推荐答案

摘自n4296草案:

8.4.2.1显式默认的函数:

明确默认的功能应

A function that is explicitly defaulted shall

(1.1)—是一个特殊的成员函数，

(1.1) — be a special member function,

(1.2)-具有相同的声明函数类型(可能的除外 不同的ref限定词，但在复制的情况下除外 构造函数或复制赋值运算符，参数类型可以是 对非常量T的引用"，其中T是成员的名称 函数的类)，就好像它已经隐式声明一样，并且

(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)-没有默认参数.

(1.3) — not have default arguments.

12.8.10复制和移动类对象:

隐式声明的类X的move构造函数将具有以下形式: X :: X(X&)

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

---

因此，该行:

---

As a consequence the line:

T(const T&&) = default;

无效，因为隐式声明的move构造函数的形式为:

is not valid because the implicitly-declared move constructor has the form of:

T(T&&)

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