c++ - 如何为模板类的 const ref 成员定义 move 赋值运算符

Question

我有以下模板类，其中成员是 const ref 类型。对象的复制被禁用，并且只希望有 move cntor 和 move 赋值运算符。

Q1:如何正确实现const ref type的 move 赋值运算符(是否正确，我做的)？

Q2:为什么会这样

MyClass<int> obj2(std::move(obj));   // will work with move ctor
MyClass<int> obj3 = std::move(obj2); // also move ctor called: Why?

发生了什么？

Q3:在 main() 中 move 的实例可以使用 print() 调用。是UB吗？

我正在使用 Visual Studio 2015 (v140)。这是我的代码:

#include <utility>
#include <iostream>

template<typename Type>
class MyClass
{
    const Type& m_ref;  // const ref type
public:
    explicit MyClass(const Type& arg): m_ref(std::move(arg)){}

    // coping is not allowed
    MyClass(const MyClass&) = delete;
    MyClass& operator=(const MyClass&) = delete;

    // enables move semantics
    MyClass(MyClass &&other) : m_ref(std::move(other.m_ref)) { std::cout << "Move Cotr...\n"; } // works

    // how would I do the move assignment operator, properly: following?
    MyClass& operator=(MyClass &&other)
    {
        // this should have been done in initilizer list(due to const ref member), 
        // but here we cannnot and still it gives no errors, why?

        this->m_ref = std::move(other.m_ref);  
        std::cout << "Move =operator...\n";
        return *this;
    }

    // print the member
    const void print()const noexcept { std::cout << m_ref << std::endl; }
};

//test program
int main() {
    MyClass<int> obj(2);
    MyClass<int> obj2(std::move(obj));   // will work with move ctor
    MyClass<int> obj3 = std::move(obj2); // also move ctor called: Why?

    obj.print();  // why this prints 2? : is it UB?
    obj2.print(); // why this prints 2? : is it UB?
    obj3.print(); // here it makes sence.

    std::cin.get();
}

Answer 1

第一个:

MyClass<int> obj2(std::move(obj));   // will work with move ctor

是direct initialization .

第二个:

MyClass<int> obj3 = std::move(obj2); // also move ctor called: Why?

是copy initialization .

两者都在构造对象(分别是obj2 和obj3)并初始化它们。 = 在此上下文中并不表示赋值。