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.
