C++ 如何在c+中通过基类引用访问派生类的对象+；？

我会用C++阅读以下几行：完整的参考书。我没有弄清楚

References to Derived Types

Similar to the situation as described for pointers earlier, a base class reference can be
used to refer to an object of a derived class. The most common application of this is
found in function parameters. A base class reference parameter can receive objects of
the base class as well as any other type derived from that base.

我对以下代码有疑问：

#include<iostream>

using namespace std;


class base
{
protected:
    int i;
public:
    void seti(int num){i=num;}
    int geti(){return i;}
};

class derived:public base
{
protected:
    int j;
public:
    void setj(int num){j=num;}
    int getj(){return j;}
};


void refe(base &ob)
{
    ob.seti(1);
    ob.setj(2);
}

int main()
{
   derived d;
   refe(d);

   cout<<d.geti();
   cout<<"\n"<<d.getj();

   return 0;
}

在函数refe（）（（派生*）ob.setj（2）中，我也像这样强制转换；这还显示以下错误：

D:\Users\srilakshmikanthan.p\Documents\source code\ex.cpp||In function 'void refe(base&)':|
D:\Users\srilakshmikanthan.p\Documents\source code\ex.cpp|28|error: 'class base' has no member named 
'setj'
=== Build failed: 1 error(s), 0 warning(s) (0 minute(s), 5 second(s)) ===

D:\Users\srilakshmikanthan.p\Documents\source code\ex.cpp||In function 'void refe(base&)':|
D:\Users\srilakshmikanthan.p\Documents\source code\ex.cpp|28|error: invalid cast from type 'base' to 
type 'derived*'
=== Build failed: 1 error(s), 0 warning(s) (0 minute(s), 3 second(s)) ===|

---

所以请在那本书中解释这几行。

您只能从基类引用中调用派生类的虚函数。请看这个

#include<iostream>

class Base
{
public:
    virtual void fun()
    {
        std::cout<<"Base";
    }
};

class Derived:public Base
{
public:
    void fun()override
    {
        std::cout<<"Derived";
    }
};

void myfun(Base &ob)
{
    ob.fun();
}

int main()
{
    Derived ob;
    myfun(ob);
    return 0;
}

在函数refe（）（（派生*）ob.setj（2）中，我也像这样强制转换；这还显示以下错误：

D:\Users\srilakshmikanthan.p\Documents\source code\ex.cpp||In function 'void refe(base&)':|
D:\Users\srilakshmikanthan.p\Documents\source code\ex.cpp|28|error: 'class base' has no member named 
'setj'
=== Build failed: 1 error(s), 0 warning(s) (0 minute(s), 5 second(s)) ===

D:\Users\srilakshmikanthan.p\Documents\source code\ex.cpp||In function 'void refe(base&)':|
D:\Users\srilakshmikanthan.p\Documents\source code\ex.cpp|28|error: invalid cast from type 'base' to 
type 'derived*'
=== Build failed: 1 error(s), 0 warning(s) (0 minute(s), 3 second(s)) ===|

---

转换为

（（派生和）ob）.setj（2）这将起作用，因为它不是指针，而是参考
要使多态性发挥作用，您需要使用
虚拟
函数。在子类中重写这些函数，而不是创建新函数。至于cast，首先，C风格的cast应该被看作是一个警告，表明您做错了什么。其次，在
refe
函数中，
ob
是引用而不是指针，因此需要将其转换为引用（如
static\u cast（ob）
），这意味着对派生类的引用可以隐式转换为对基类的引用。这并不意味着您可以使用对基类的引用来访问派生类的成员。要通过基类引用调用派生行为，必须使用
virtual
语义，或者将引用转换回对派生类的引用。