class base
{
public:
void virtual func(){cout<<"base";}
void check()
{
func();
}
};
class derived: public base
{
public:
void func(){cout<<"dervied";}
};
int main()
{
base *obj = new derived();
obj->check();
return 0;
}
Above code prints derived on the console.
Now, I understand the concept of virtual functions but I'm unable to apply it here. In my understanding whenever we call a virtual function, compiler modifies the call to "this->vptr->virtualfunc()" and that's how most heavily derived's class function gets invoked. But in this case, since check() is not a virtual function, how does the compiler determine that it needs to invoke func() of derived?
how does the compiler determine that it needs to invoke func() of derived?
In the same exat way - by invoking this->vptr->virtualfunc(). Recall that this belongs to the derived class even inside the base class, because each derived class is a base class as well, so the same way of accessing virtual functions works for it too.
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