友元函数与继承

在基类中定义了友元函数，在派生类继承了基类中的友元函数吗？

用例子来说明：

#include <iostream>
using namespace std;
class Increase
{
public:
    Increase(int x):value(x){}

    friend Increase& operator++(Increase&); //前增量
    friend Increase operator++(Increase&, int); //后增量

    void display(){ cout << "the value of Increase is " << value <<endl;}

protected:
    int value;
};

Increase& operator++(Increase& a)
{
    a.value++;
    return a;
}
Increase operator++(Increase& a, int)
{
    Increase temp(a);
    a.value++;
    return temp;
}

class Derived: public Increase
{
public:
    Derived(int x):Increase(x){}
};

int main()
{
    Derived der(10);
    der.display();
    (++der).display();

    return 0;
}

运行的结果为：

the value of Increase is  10

the value of Increase is  11

从这个结果来看：似乎可以得出“派生类继承了基类的友元函数”的结论。但是，我们将代码稍微

修改一下：

#include <iostream>
using namespace std;
class Increase
{
public:
    Increase(int x):value(x){}

    friend   Increase& operator++(Increase&);           //前增量
    friend   Increase operator++(Increase&, int);         //后增量

    void   display(){ cout << "the value of Increase is " << value <<endl;}

protected:
    int   value;
};

Increase& operator++(Increase& a)
{
    a.value++;
    return   a;
}
Increase operator++(Increase& a, int)
{
    Increase   temp(a);
    a.value++;
    return   temp;
}

class Derived: public Increase
{
public:
    Derived(int x):Increase(x){}

    friend Derived& operator++(Derived&); //前增量
    void display(){ cout << "the value of Derived is " << value <<endl; }
};
Derived& operator++(Derived& a)
{
    a.value++;
    a.value++;
}

int main()
{
    Derived   der(10);
    der.display();
    (++der).display();

    return 0;
}

运行结果为：

the value of Derived is  10

the value of Derived is  12

可以看出，派生类根本没有继承基类的友元函数，但是我们可以将派生类的对象传递给基类的友元函数，

他会自动转换成基类的对象，使得基类的友元函数也可以调用成功。