为什么operator =()函数不能由derive类继承

Question

可能重复:
在C++中继承operator =的麻烦

我更新了代码

#include <QtCore/QCoreApplication>

class Base
{
    int num;
public:
    Base& operator=(int rhs)
    {
        this->num = rhs;
        return *this;
    }
};

class Derive : public Base
{
public:
    int deriveNum;
    using Base::operator =; // unhide the function
};

int main(int argc, char *argv[])
{
    QCoreApplication a(argc, argv);

    Base base;
    Derive derive1, derive2;

    base = 1;  // calls Base::operator(1) and returns Base&

    derive1 = 11; // calls Base::operator(11) and returns Base&
    derive2 = 22; // calls Base::operator(22) and returns Base&

    derive1 = base;// Which function does it calls??
               // If it calls Base::operator(base) and
                   // returns a Base&, how could it be assigend to derive1?

    return a.exec();
}

我在评论中标出了这个问题,请给我一些帮助

Answer 1

它是由派生类继承.但是,派生类具有它自己operator =(由编译器隐含声明),它隐藏了operator =从父类继承的内容(在C++中搜索和读取"名称隐藏").

如果您希望继承operator =变为可见,则必须明确取消隐藏它

class Derive : public Base
{
  std::string str;

public:
  using Base::operator =; // unhide
};

你的代码将编译.(如果您修复了明显的语法错误.请发布实际代码.)

PS这个问题经常被问到.我提供了一个更详细解释的链接,作为对您问题的评论.