c++ - 为什么对象可以是 "moved"甚至缺少 move 构造函数和 move 赋值运算符？

Question

http://en.cppreference.com/w/cpp/language/rule_of_three

几个月前我开始使用 c++11并观看了五人规则。

所以..我开始将复制构造函数/复制赋值运算符/move 构造函数/move 赋值运算符与 default 关键字放在每个具有虚拟析构函数的类上。

因为规则告诉我，如果您声明显式析构函数，那么您的类将不再具有隐式 move 构造函数和 move 赋值运算符。

所以我认为 gcc 会因为缺少 move 构造函数和 move 赋值运算符而向我提示下面的类。

但是效果很好!发生了什么？？

class Interface {
public:
    virtual ~Interface() = default;  // implicit destructor
};

class ImplA : public Interface {
public:
    virtual ~ImplA() = default; // implicit destructor
};

ImplA first;
ImplA second(first);            // copy constructor, OK. understood it.
ImplA third(std::move(first));  // move constructor, it's OK. Why?
second = first;             // copy assignment, OK. understood it.
second = std::move(first);  // move assignment, it's also OK. Why?

Answer 1

So I think gcc is going to complain to me that below class due to lack of move constructor and move assignment operator.

因为可以通过复制构造函数和复制赋值运算符执行所需的操作。 Interface 仍然有复制构造函数和复制赋值运算符，它们是 implcitly declared .右值总是可以绑定(bind)到 const Interface&。

更准确地说，即使没有提供 move 构造函数和 move 赋值运算符，Interface 仍然是 MoveConstructible ,

A class does not have to implement a move constructor to satisfy this type requirement: a copy constructor that takes a const T& argument can bind rvalue expressions.

和MoveAssignable .

The type does not have to implement move assignment operator in order to satisfy this type requirement: a copy assignment operator that takes its parameter by value or as a const Type&, will bind to rvalue argument.

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顺便说一句:如果您创建 move 构造函数和 move 赋值运算符 delete明确地那么复制和 move 操作都会失败。与右值表达式一起使用，将选择显式删除的重载，然后失败。使用左值表达式也会失败，因为复制构造函数和复制赋值运算符由于 move 构造函数或 move 赋值运算符的声明而被隐式声明为已删除。