为什么通过指针调用fu1()\fn2()函数不能输出值？代码应该怎么改？

Question

//8-8.cpp

/*定义一个基类BaseClass，从它派生出类DerivedClass.
BaseClass有成员函数fn1()(虚函数),fn2().
DerivedClass也有成员函数fn1(),fn2()
在主函数中声明一个派生类对象，分别用基类和派生类的指针指向派生类对象，
并通过指针调用两个函数*/
#include <iostream>
#include "classfile.h"
using namespace std;
int fun(BaseClass* ptr)
{
    ptr->fn1();
}
int fun2(BaseClass* ptr)
{
    ptr->fn2();
}
int main()
{
    BaseClass b1(1, 2);
    DerivedClass d1(1, 2);
    cout << fun(&b1) << endl;
    cout << fun(&d1) << endl;
    cout << fun2(&b1) << endl;
    cout << fun2(&d1) << endl;
    return 0;
}

//classfile.h

#pragma once
#include <iostream>
using namespace std;
class BaseClass {
private:
    int x, y;
public:
    BaseClass(int x=0,int y=0):x(x),y(y){}
    ~BaseClass(){}
    virtual int fn1()const;  //fn1()为虚函数
    int fn2()const;
};
class DerivedClass:public BaseClass {
public:
    DerivedClass(int x, int y);
    int fn1()const;
    int fn2()const;
};

//classfile.cpp

#include <iostream>
#include "classfile.h"
using namespace std;
int BaseClass::fn1()const
{
    return x;
}
int BaseClass::fn2()const
{
    return y;
}
int DerivedClass::fn1()const
{
    return  (BaseClass::fn1()+1);
}
int DerivedClass::fn2()const
{
    return BaseClass::fn2() + 1;
}

Answer 1

int fun(BaseClass* ptr)
{
    ptr->fn1();
}
->
int fun(BaseClass* ptr)
{
    return ptr->fn1();
}