About references and pointers on classes...

Question

Why the output of the following first code is 3, while the second is 1?
#include<iostream>
using namespace std;
class A{
public:
virtual void f() {cout<<1;}
};
class B:public A{
public:
void f(){cout<<2;}
};
class C:public B{
public:
void f(){cout<<3;}
};
int main()
{
B *p=new C;
p->f();
}

################

#include <iostream>
using namespace std;
class A{
public:
void f(){cout<<1;}
};
class B: public A{
private:
void f(){cout<<2;}
};
void g(A& a){a.f();}
int main()
{
B b;
g(b);
return 0;
}

Answer 1

It has nothing to do with pointer and reference, it is because of the virtual keyword that you have 3 in first input (you should look once again at the course if you do not get it) and because B overloaded function is private that you have 1 instead of 2 in the second case

Answer 2

But it seems "private or not" cannot explain the second circumstance, as I've executed the following codes, of which the output was still one. I think it has something to do with some concepts of "reference".
#include <iostream>
using namespace std;
class A{
public:
void f(){cout<<1;}
};
class B: public A{
public:
//private:
void f(){cout<<2;}
};
void g(A& a){a.f();}
int main()
{
B b;
g(b);
return 0;
}

Answer 3

Yes as a is of type A and f is not virtual

with the private you removed, directly doing
b.f(); instead of g(b); would have had the same result

Answer 4

Very clear and detailed explanation, thanks so much.:)

Answer 5

But my interpreter seems not to explain in that way. When directly doing b.f(), I got an error saying "'void B::f()' is private within this context...", rather than using the f() function in base class...

Answer 6

Then it is my mistake ! You just can't :)