当派生类和基类的虚构函数默认实参不同时的传递问题
当派生类和基类的虚构函数默认实参不同时；
派生类的实参将接收基类传递过来的默认参数
示列如下：
[cpp]
#include <iostream>
#include <ostream>
#include <string>
using namespace std;
class B{
public:
virtual void fun(int num = 0)
{
cout << "B = " << num << endl;
}
};
class D: public B{
public:
virtual void fun(int num = 1)
{
cout << "D = " << num << endl;
}
};
int main()
{
D d;
B *p = &d;
p -> fun();
B &b = d;
b.fun();
return 0;
}
#include <iostream>
#include <ostream>
#include <string>
using namespace std;
class B{
public:
virtual void fun(int num = 0)
{
cout << "B = " << num << endl;
}
};
class D: public B{
public:
virtual void fun(int num = 1)
{
cout << "D = " << num << endl;
}
};
int main()
{
D d;
B *p = &d;
p -> fun();
B &b = d;
b.fun();
return 0;
}
运行结果如下：
D = 0
D = 0
补充：软件开发 , C++ ,
