c – 是一个私人移动构造函数来防止移动？

C中的常见模式是使副本构造函数为private：

class A
{
    public:
        // ...
    private:
        A(const A&);
};

但是下面的代码将会编译(在C 11/14中)：

A f();

auto a = f();

该标准包含有关自动生成移动构造函数的信息.我既没有访问标准也没有实际生成移动构造函数的编译器.我的问题是：我必须写

class A
{
    public:
        // ...
    private:
        A(const A&);
        A(const A&&);
};

以防止移动(和类似运算符)？

解决方法

But will the following code then compile (in C++11/14):

不,不会.用户声明的复制构造函数的存在应该阻止移动构造函数的隐式生成.根据C11标准第12.8 / 9段：

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.