为什么既没有用clang声明也没有删除move构造函数? [英] Why is the move constructor neither declared nor deleted with clang?

问题描述

请考虑以下课程.

struct with_copy {
    with_copy() = default;
    with_copy(with_copy const&) {}
    with_copy& operator=(with_copy const&) { return *this; }
};

struct foo {
    with_copy c;
    std::unique_ptr<int> p;
};

• with_copy是否具有副本构造函数?是的.它是明确定义的.
• with_copy是否具有移动构造函数?否.显式复制构造函数阻止生成它.
• with_copy是否具有已删除的move构造函数?不.没有移动构造函数与拥有被删除的构造函数不同.删除的move构造函数将尝试移动格式不正确的文件，而不是退化为副本.
• with_copy是否可复制?是的.其副本构造函数用于副本.
• with_copy是否可移动?是的.其副本构造函数用于移动.
• Does with_copy have a copy constructor? Yes. It was explicitly defined.
• Does with_copy have a move constructor? No. The explicit copy constructor prevents it from being generated.
• Does with_copy have a deleted move constructor? No. Not having a move constructor is not the same as having a deleted one. A deleted move constructor would make an attempt to move ill-formed instead of degenerating to a copy.
• Is with_copy copyable? Yes. Its copy constructor is used for copies.
• Is with_copy movable? Yes. Its copy constructor is used for moves.

...现在是棘手的人.

... and now the tricky ones.

• foo是否具有副本构造函数?是的.它已删除了一个，因为调用unique_ptr的已删除副本构造函数，其默认定义将不正确.
• foo是否具有move构造函数? GCC同意，c拒绝.
• foo是否具有已删除的move构造函数? GCC和clang都拒绝.
• foo是否可复制?否.其副本构造函数已删除.
• foo是否可移动? GCC同意，c拒绝.
• Does foo have a copy constructor? Yes. It has a deleted one, as its defaulted definition would be ill-formed due to invoking unique_ptr's deleted copy constructor.
• Does foo have a move constructor? GCC says yes, clang says no.
• Does foo have a deleted move constructor? Both GCC and clang say no.
• Is foo copyable? No. Its copy constructor is deleted.
• Is foo movable? GCC says yes, clang says no.

(当人们考虑分配而不是构造时，行为类似.)

(The behaviour is similar when one considers assignment instead of construction.)

据我所知，GCC是正确的. foo应该具有对每个成员执行移动的move构造函数，在with_copy的情况下，该成员会退化为副本. Clang的行为似乎很荒谬:我有一个带有两个可移动成员的聚合，但是我的聚合是一个不可移动的砖.

As far as I can see, GCC is correct. foo should have a move constructor that performs a move on each member, which in with_copy's case degenerates to a copy. Clang's behaviour seems quite ridiculous: I have an aggregate with two movable members, and yet my aggregate is an immovable brick.

谁是对的?

推荐答案

C ++ 11或n3485，[class.copy]/9:

C++11, or rather n3485, [class.copy]/9:

如果类X的定义未明确声明move构造函数，则将隐式声明一个move构造函数 且仅当且仅当

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没有用户声明的副本构造函数，
• X没有用户声明的副本分配运算符，
• X没有用户声明的移动分配运算符，
• X没有用户声明的析构函数，并且
• move构造函数不会隐式定义为已删除.
• 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.

和/11:

隐式声明的复制/移动构造函数是其类的inline public成员.默认副本/ 如果X具有以下内容，则将类X的move构造函数定义为已删除(8.4.3):

An implicitly-declared copy/move constructor is an inline public member of its class. A defaulted copy/ move constructor for a class X is defined as deleted (8.4.3) if X has:

• [...]
• 对于复制构造函数，是rvalue引用类型的非静态数据成员，或者
• 对于move构造函数，是非静态数据成员或直接或虚拟基类，其类型为 没有移动构造函数，并且不可复制.
• [...]
• for the copy constructor, a non-static data member of rvalue reference type, or
• for the move constructor, a non-static data member or direct or virtual base class with a type that does not have a move constructor and is not trivially copyable.

由于with_copy是 不可复制的，所以foo将具有 no 移动构造函数(它将被定义为已删除，因此不会被隐式包含)声明).

As with_copy is not trivially copyable, foo will have no move-constructor (it would be defined as deleted, therefore it won't be implicitly declared).

C ++ 1y，或者说是github repo从2013-11-12提交e31867c0；合并 DR1402

C++1y, or rather github repo commit e31867c0 from 2013-11-12; incorporating DR1402:

/9:

如果类X的定义未明确声明移动 构造函数，当且仅当一个构造函数被隐式声明为默认构造函数 如果

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没有用户声明的副本构造函数，
• X没有用户声明的副本分配运算符，
• X没有用户声明的移动分配运算符，并且
• X没有用户声明的析构函数.
• 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.

和/11:

隐式声明的复制/移动构造函数是inline public 同类的成员.类X的默认复制/移动构造函数 如果X具有:

An implicitly-declared copy/move constructor is an inline public member of its class. A defaulted copy/ move constructor for a class X is defined as deleted (8.4.3) if X has:

• [...]
• 对于复制构造函数，是rvalue引用类型的非静态数据成员.

定义为删除的默认移动构造函数将被忽略 重载分辨率(13.3，13.4).

A defaulted move constructor that is defined as deleted is ignored by overload resolution (13.3, 13.4).

在这里，foo将具有移动构造器.

Here, foo will have a move-constructor.

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