如何从派生类访问基类中的重载运算符？

Question

见以下代码：

#include<iostream>

using namespace std;


class ex
{
    int i;
public:
    ex(int x){i=x;}
    void operator-()
    {
        i=-i;
    }
    int geti(){return i;}
};

class derived:public ex
{
    int j;
public:
    derived(int x,int y):ex(x){j=y;}
    void operator-()
    {
     j=-j;
    }
    int getj(){return j;}
};


int main()
{
    derived ob(1,2);
    -ob;
    cout<<ob.geti();
    cout<<"\n"<<ob.getj();
}

输出：

1
-2
Process returned 0 (0x0)   execution time : 0.901 s
Press any key to continue.

我在基类和派生类中都定义了- 运算符，但-ob; 只调用派生类的运算符。那么如何也将i字段改为-i（调用基类中的操作符）。

我需要任何显式函数来实现这一点吗？

Answer 1

看来你的意思

void operator-()
{
    ex::operator -();
    j=-j;
}

在任何情况下，最好像声明这样的运算符

ex & operator-()
{
    i = -i;

    return *this;
}

和

derived & operator-()
{
    ex::operator -();

    j = -j;

    return *this;
}

您也可以将操作员设为虚拟。例如

#include<iostream>

using namespace std;


class ex
{
    int i;
public:
    virtual ~ex() = default;

    ex(int x){i=x;}
    virtual ex & operator-()
    {
        i = -i;

        return *this;
    }

    int geti(){return i;}
};

class derived:public ex
{
    int j;
public:
    derived(int x,int y):ex(x){j=y;}
    derived & operator-() override
    {
        ex::operator -();

        j = -j;

        return *this;
    }
    int getj(){return j;}
};


int main()
{
    derived ob(1,2);

    ex &r = ob;

    -r;

    cout<<ob.geti();
    cout<<"\n"<<ob.getj();
}    

Answer 2

你的派生类可以这样声明：

class derived:public ex
{
    int j;
public:
    derived(int x,int y):ex(x){j=y;}
    void operator-()
    {
        j=-j;
        ex::operator-();
    }
    int getj(){return j;}
};