Visibility of members in inheritance

[ctick]

class A
{
public:
method_A() { cout << "A method " << endl; }
};

class B : public A
{
public:
method_B() { cout << "B method " << endl; }
};

int main()
{
A* aptr = new B;
....
}

Why cann't "aptr" see method_B() of class B since "aptr" is a pointer
pointing actually at an object of type B?

Thanks in advance!

[Victor Bazarov]

"ctick" <ct***@flare.com> wrote...

class A
{
public:
method_A() { cout << "A method " << endl; }
You mean

void method_A() { ...
};

class B : public A
{
public:
method_B() { cout << "B method " << endl; }
You mean

void method_B() { ...
};

int main()
{
A* aptr = new B;
...
}

Why cann't "aptr" see method_B() of class B since "aptr" is a pointer
pointing actually at an object of type B?

Because the _static_ type of aptr is A*. And it is not "pointing
actually" at a B, it is pointing to an A inside the B.

If you want to get to 'method_B', you will have to static_cast your
'aptr' to 'B*':

static_cast<B*>(aptr)->method_B();

Victor