Overrriding (interesting example)

[sawer]

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

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.

[Eli Gassert]

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.

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

[Eli Gassert]

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

[sawer]

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