赋值运算符在派生类中不可用 [英] Assignment operator not available in derived class

问题描述

基类中的赋值运算符在派生类中似乎不可用.鉴于此代码:

The assignment operator in base class does not seem to be available in derived class. Given this code:

#include <iostream>

class A{
    int value;
public:
    A& operator=(int value){
        this->value = value;
        return *this;
    }
};

class B : public A{};

int main(){
    B b;
    b = 0; // Does not work
    return 0;
}

GCC 6.4 说:

错误:'operator=' 不匹配(操作数类型为 'B' 和 'int')

error: no match for 'operator=' (operand types are 'B' and 'int')

发生了什么?

推荐答案

当我们自己不提供赋值运算符时，每个类至少有一个隐式定义的赋值运算符.

Every class has at least one assignment operator implicitly defined when we don't provide one ourselves.

当派生类中的成员函数定义为与基类中的成员同名时，它会隐藏该名称的所有基类定义.

And when a member function in a derived class is defined with the same name as a member in the base class, it hides all the base class definitions for that name.

您可以使用 using 声明，但要注意它会拉所有名为 operator= 的成员并允许这样的代码:

You can use a using declaration, but be warned that it will pull all the members named operator= and allow code like this:

A a;
B b;
b = a;

这有点可疑.

这篇关于赋值运算符在派生类中不可用的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！