c++ - C++中移动赋值运算符的继承

Question

我需要一些帮助来理解移动赋值运算符的继承过程。 对于给定的基类

class Base
{
public:
    /* Constructors and other utilities */
    /* ... */

    /* Default move assignment operator: */
    Base &operator=(Base &&) = default;
    /* One can use this definition, as well: */
    Base &operator=(Base &&rhs) {std::move(rhs); return *this;}

    /* Data members in Base */
    /* ... */
};

class Derived : public Base
{
public:
    /* Constructors that include inheritance and other utilities */
    /* ... */

    Derived &operator=(Derived &&rhs);

    /* Additional data members in Derived */
    /* ... */
};

我不太确定如何在派生类中调用基本移动赋值运算符？我应该只使用范围运算符并说

Base::std:move(rhs);

后面是 Derived 类中定义的附加项的后续 std::move(...)，还是有其他方法？

Answer 1

要调用继承的operator=，您通常会调用继承的operator=。

Derived &operator=(Derived &&rhs) {
   Base::operator=(std::move(rhs));
   // do the derived part
   return *this;
}

无论是复制赋值、移动赋值还是某种用户定义的赋值，模式都是相同的。