class A{
public:
// data members
virtual void serialize()
{
// code
}
};
class B : public A
{
public:
void serialize()
{
// code
}
};
int main()
{
B bObject;
// call serialize for class A and B here
return 0;
}
Accessing base class functions
Author
Is it possible, when dealing with an derived object, to call the functions for the base class?
Because I'm a little lazy and have a base class with about 17 data members, is it possible to call A::serialize and then B::serialize, both on the B object? I mean, I doubt it, because of, you know, typing and all, but, good to check anyway.
Might you be looking for...
bObject.Serialise();
dynamic_cast<A>(bObject).Serialise();
? You should even be able to get away with a static_cast<A>, depending on whether B is inherited from A, which in your example it is... So yeah, static_cast should work...
bObject.Serialise()
static_cast<A>(bObject).Serialise();
Or am I missing your point here?
CJM
bObject.Serialise();
dynamic_cast<A>(bObject).Serialise();
? You should even be able to get away with a static_cast<A>, depending on whether B is inherited from A, which in your example it is... So yeah, static_cast should work...
bObject.Serialise()
static_cast<A>(bObject).Serialise();
Or am I missing your point here?
CJM
Sure, just specify the class for which you want to call the method like this:
-Markus-
class Base { public: virtual void foo() {}};class Derived : public Base { public: virtual void foo() {}};int main() { Derived x; x.Base::foo(); x.Derived::foo();}-Markus-
...or did you intend to automatically call A::serialize() upon B::serialize() being called. That's the responsibility of B's implementation, meaning you should do it like this
-Markus-
//// B's serialize() implementation//void B::serialize() { A::serialize(); // Serialize own stuff}-Markus-
This topic is closed to new replies.
Advertisement
Popular Topics
Advertisement