C++11 move 构造函数

Question

考虑到以下类，实现 move 构造函数的正确方法是什么:

class C {
public:
    C();
    C(C&& c);
private:
    std::string string;
}

当然，这样做的目的是避免复制 string 或将其释放两次。
让我们假设基本示例只是为了清楚起见，我确实需要一个 move 构造函数。

我试过了:

C::C(C&& c) {
    //move ctor
    string = std::move(c.string);
}

和

C::C(C&& c) : string(std::move(c.string)) {
    //move ctor
}

两者都可以在 gcc 4.8 上编译并运行良好。 似乎选项 A 是正确的行为，string 被复制而不是与选项 B 一起 move 。
这是 move 构造函数的正确实现吗？

Answer 1

由于 std::string 本身有一个 move-ctor，因此为 C 隐式定义的 move-ctor 将负责正确的 move 操作。您可能无法自己定义。但是，如果您有任何其他数据成员，特别是:

12.8 Copying and moving class objects

12 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:

— a variant member with a non-trivial corresponding constructor and X is a union-like class,

— a non-static data member of class type M (or array thereof) that cannot be copied/moved because overload resolution (13.3), as applied to M’s corresponding constructor, results in an ambiguity or a function that is deleted or inaccessible from the defaulted constructor, or

— a direct or virtual base class B that cannot be copied/moved because overload resolution (13.3), as applied to B’s corresponding constructor, results in an ambiguity or a function that is deleted or inaccessible from the defaulted constructor, 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.

13 A copy/move constructor for class X is trivial if it is neither user-provided nor deleted and if

— class X has no virtual functions (10.3) and no virtual base classes (10.1), and functions (10.3) and no virtual base classes (10.1), and

— the constructor selected to copy/move each direct base class subobject is trivial, and

— for each non-static data member of X that is of class type (or array thereof), the constructor selected to copy/move that member is trivial; otherwise the copy/move constructor is non-trivial.

您可能想要实现自己的 move-ctor。

如果您需要 move-ctor，请首选初始化列表语法。总是!否则，您最终可能会得到初始化列表中未提及的每个对象的默认构造(对于具有非默认 ctors 的成员对象，您必须这样做)。