On Apr 19, 1:16 am, Venkat <swara...@yahoo.com> wrote:
Hi,
Is there a good explanation why compilation tags error for the call to
AA.method(). It is public in the base class. If cast to baseclass,
then compiler says it's ok.
Its not ok.
Casting to baseclass calls the public A::method(), not the private
AA::method().
When calling an overridden virtual member function polymorphically,
Only the access specifier in the pointer's type is relevent.
When calling the member function directly,
the access specifier in the type involved is relevent, as expected.
So the reason is: the interface(s) provided dictates a member
function's accessibility.
The pointer to baseclass does not have the same interface as the
derived object does in this case.
Why? because it was written that way.
see void accessor() below...
class A {
public:
A() {}
virtual ~A() {}
virtual void method(void) { std::cout << "A::method" <<
std::endl; }
};// A
class AA : public A {
public:
AA() {}
virtual ~AA() {}
private:
virtual void method(void) {
A::method();
std::cout << "AA::method" << std::endl;
}
public:
void accessor()
{
method();
}
};// A
int main()
{
AA obj;
A* p = &obj;
p->method();
//((A&)obj).method();
// calls A::method() on the object's base
static_cast< A >(obj).method();
// now has a public interface to access the private virtual
override
obj.accessor();
obj.method(); // ERROR:
Its an error because you chose that interface.
class(es). If so, method() is visible via the base class A. or
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