Java中继承、多态、重载和重写介绍(5)
例6
#include <iostream>
using namespace std;
class Base{
public:
virtual void fun() { cout << "Base::fun()" << endl; }//overload
virtual void fun(int i) { cout << "Base::fun(int i)" << endl; }//overload
};
class Derive : public Base{
public:
void fun() { cout << "Derive::fun()" << endl; }//override
void fun(int i) { cout << "Derive::fun(int i)" << endl; }//override
void fun(int i,int j){ cout<< "Derive::fun(int i,int j)" <<endl;}//overload
};
int main()
{
Base *pb = new Derive();
pb->fun();
pb->fun(1);
//下面一句错误,故屏蔽掉
//pb->fun(1,2);virtual函数不能进行overload,error C2661: 'fun' : no overloaded function takes 2 parameters
cout << endl;
Derive *pd = new Derive();
pd->fun();
pd->fun(1);
pd->fun(1,2);//overload
delete pb;
delete pd;
return 0;
}
/*
输出结果
Derive::fun()
Derive::fun(int i)
Derive::fun()
Derive::fun(int i)
Derive::fun(int i,int j)
Press any key to continue
*/
例7-1
#include <iostream>
using namespace std;
class Base{
public:
virtual void fun(int i){ cout <<"Base::fun(int i)"<< endl; }
};
class Derive : public Base{};
int main()
{
Base *pb = new Derive();
pb->fun(1);//Base::fun(int i)
delete pb;
return 0;
}
例7-2
#include <iostream>
using namespace std;
class Base{
public:
virtual void fun(int i){ cout <<"Base::fun(int i)"<< endl; }
};
class Derive : public Base{
public:
void fun(double d){ cout <<"Derive::fun(double d)"<< endl; }
};
int main()
{
Base *pb = new Derive();
pb->fun(1);//Base::fun(int i)
pb->fun((double)0.01);//Base::fun(int i)
delete pb;
return 0;
}
例8-1
#include <iostream>
using namespace std;
class Base{
public:
virtual void fun(int i){ cout <<"Base::fun(int i)"<< endl; }
};
class Derive : public Base{
public:
void fun(int i){ cout <<"Derive::fun(int i)"<< endl; }
};
int main()
{
Base *pb = new Derive();
pb->fun(1);//Derive::fun(int i)
delete pb;
return 0;
}
例8-2
#include <iostream>
using namespace std;
class Base{
public:
virtual void fun(int i){ cout <<"Base::fun(int i)"<< endl; }
};
class Derive : public Base{
public:
void fun(int i){ cout <<"Derive::fun(int i)"<< endl; }
void fun(double d){ cout <<"Derive::fun(double d)"<< endl; }
};
int main()
{
Base *pb = new Derive();
pb->fun(1);//Derive::fun(int i)
pb->fun((double)0.01);//Derive::fun(int i)
delete pb;
return 0;
}
例9
#include <iostream>
using namespace std;
class Base{
public:
virtual void fun(int i){ cout <<"Base::fun(int i)"<< endl; }
};
class Derive : public Base{
public:
void fun(int i){ cout <<"Derive::fun(int i)"<< endl; }
void fun(char c){ cout <<"Derive::fun(char c)"<< endl; }
void fun(double d){ cout <<"Derive::fun(double d)"<< endl; }
};
int main()
{
Base *pb = new Derive();
pb->fun(1);//Derive::fun(int i)
pb->fun('a');//Derive::fun(int i)
pb->fun((double)0.01);//Derive::fun(int i)
Derive *pd =new Derive();
pd->fun(1);//Derive::fun(int i)
//overload
pd->fun('a');//Derive::fun(char c)
//overload
pd->fun(0.01);//Derive::fun(double d)
delete pb;
delete pd;
return 0;
}
例7-1和例8-1很好理解,我把这两个例子放在这里,是让大家作一个比较摆了,也是为了帮助大家更好的理解:
n 例7-1中,派生类没有覆盖基类的虚函数,此时派生类的vtable中的函数指针指向的地址就是基类的虚函数地址。
n 例8-1中,派生类覆盖了基类的虚函数,此时派生类的vtable中的函数指针指向的地址就是派生类自己的重写的虚函数地址。
在例7-2和8-2看起来有点怪怪,其实,你按照上面的原则对比一下,答案也是明朗的:
n 例7-2中,我们为派生类重载了一个函数版本:void fun(double d) 其实,这只是一个障眼法。我们具体来分析一下,基类共有几个函数,派生类共有几个函数:
类型
基类
派生类
Vtable部分
void fun(int i)
指向基类版的虚函数void fun(int i)
静态部分
void fun(double d)
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