--- In [email protected], saurabh jain <hundredr...@...> wrote:
>
> Hi,
> Can anyone tell what is supposed to be the behaviour for a function which
> is declared as virtual in a base class and non-virtual in derived?
> Specially pertaining to cases when you cast the derived object into a base
> class object and call the function.
>
> For example:
>
> class Base {
> public:
> virtual void foo() { cout << " I am Base foo \n"; }
> };
>
> class Derived : public Base {
> public:
> void foo() { cout << " I am Derived foo\n";}
> };
>
For my first question this is what I understood from the standard:
"If a virtual member function vf is declared in a class Base and in a class
Derived, derived directly or
indirectly from Base, a member function vf with the same name and same
parameter list as Base::vf is
declared, then Derived::vf is also virtual (whether or not it is so declared)
and it overrides95)
__________________
95) A function with the same name but a different parameter list (13) as a
virtual function is not necessarily virtual and does not over10.3
Virtual functions Derived classes 10 7
Base::vf."
> And to extend the question what if another class Derived2 is derived from
> class "Derived" with the function foo being declared as non-virtual.
>
For the second question :
"[Note: a virtual member function does not have to be visible to be overridden,
for example,
struct B {
virtual void f();
};
struct D : B {
void f(int);
};
struct D2 : D {
void f();
};
the function f(int) in class D hides the virtual function f() in its base class
B; D::f(int) is not a virtual
function. However, f() declared in class D2 has the same name and the same
parameter list as
B::f(), and therefore is a virtual function that overrides the function B::f()
even though B::f() is
not visible in class D2. ]
"
> Thanks and Regards,
> Saurabh
>
> $ make war
> make: *** No rule to make target `war'. Try `peace' instead
>
