派生类继承基类的赋值运算符？

问题描述

在我看来，Derived类不继承基类赋值运算符

如果Derived类继承Base类赋值运算符，请解释下面的示例

It seems to me that Derived class don't inherit base class Assignment operator
if Derived class inherit Base class assignment operator , can you please explain the following example

在下面的代码中我覆盖了基类operator =在Derived中，所以Derived类默认赋值操作符调用重载operator =

In the following code I am overriding base class operator= in Derived, so that Derived class default assignment operator calls overloaded operator=

#include <iostream>
using namespace std;
class Base
{
    public:
    Base(int lx = 0):x(lx)
    {
    }

    virtual Base& operator=( const Base &rhs)
    {
        cout << "calling Assignment operator in Base" << endl;
        return *this;
    }

    private:
    int x;
};


class Derived : public Base
{
    public:
    Derived(int lx, int ly): Base(lx),y(ly)
    {
    }

    Base& operator=(const Base &rhs)
    {
        cout << "Assignment operator in Derived"<< endl;
        return *this;
    }

    private:
    int y;
};



int main()
{
    Derived d1(10,20);
    Derived d2(30,40);
    d1 = d2;
}

它提供输出

我已将基类operator =重写为派生类，所以如果派生类继承基类operator =，那么它应该被operator =（我已经写在派生类中）覆盖，现在Derived类default operator =应该调用覆盖版本，而不是从基类operator =。

I have re-written base class operator= into derived class, so if derived class inherits base class operator= then it should be get overridden by operator= (that i have written in derived class), and now Derived class default operator= should call overridden version and not from the base class operator=.

推荐答案

编译器为Derived生成一个默认赋值运算符但是，默认赋值运算符会调用类的成员和基类的所有赋值运算符。

The compiler generates a default assignment operator for Derived (which hides the operator of Base). However, the default assignment operator calls all assignment operators of the class' members and base classes.

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