Overrriding (interesting example)

I see an article about this issue. There is a code:

Code:

---

class A
{
public: 
    virtual void foo(int n = 3) 
    {   
        std::cout << "A::foo(" << n << ")" << endl; 
    }
};
class B : public A
{
public: 
    virtual void foo() 
    {    std::cout << "B::foo" << endl; 
    }
};

int main()
{ 
    A* a = new B;  a->foo(); // prints A::foo("3");
}

---

Also i tried without virtual keyword, it also gives same result. But i don't understand why it is so.

Code:

---

class A
{
public: 
    void foo(int n = 3) 
    {   
        std::cout << "A::foo(" << n << ")" << endl; 
    }
};
class B : public A
{
public: 
    void foo() 
    {    std::cout << "B::foo" << endl; 
    }
};

int main()
{ 
    A* a = new B;  a->foo();
        A as;
        as.foo();

}

---

How can A see the foo() function?
I can call as.foo() not as.foo(3); it doesnt give me error but it prints A::foo("3") but i didnt call a.foo(int n = 3);

What does compiler do for such a situation when i inherite a class.
A can see foo() function; is this normal? Because i know that function name resolution is only legal for scope. So A mustnt see foo() function. But it sees.

Can anyone please explain this.

Thanks.

This is almost word-for-word for how I see this example played out in a few of the "in depth" series books by A&W :)

You need to use the using statement.

Quote:

---

class B : public A
{
public:
using A::foo;
void foo()
{ std::cout << "B::foo" << endl;
}
};

---

I found an example of how the full situation plays out:
http://publib.boulder.ibm.com/infoce...se_derived.htm

Ahh i made a silly mistake.
as doesnt call B's function it also calls void foo(int n = 3). Because its parameter is int n = 3 not int n.
Sorry and thanks Eli Gassert