Initialize a object with a derived class constructor - c++

Question

Consider the following C++ code:
#include <iostream>
using std::cout;
class A
{
public:
int a;
A():a(0)
{
cout << "A constructor\n";
}
virtual void f()
{
cout << "f inside A\n";
}
};
class C : public A
{
public:
int c;
virtual void f()
{
cout << "f inside C\n";
}
C():c(0)
{
cout << "C constructor\n";
}
};
int main()
{
A varA = C();
cout << "Size of C class: " << sizeof(C) << "\n";
cout << "Size of varA object: " << sizeof(varA) << "\n";
C* varC = static_cast<C*>(&varA);
varC->f();
cout << "varC->a is " << varC->a << "\n";
cout << "varC->c is " << varC->c << "\n";
}
The output of this program is:
A constructor
C constructor
Size of C class: 16
Size of varA object: 8
f inside A
varC->a is 0
varC->c is 1726166356
I initialize the varA object with the constructor of class C. Constructors of A and C class are called, but the varA is a simply a A object. I cast the address of varA to C* type and I try to call its f() function, but it prints the f() function of class A, so I deduce that it is use the early binding mechanism to call it.
I think if I call the constructor of derived class, like this case, I obtain the same object if I had called the base constructor.
I think the only difference is the other constructors are called. Is my supposition right or there are any other differences?

Answer 1

Classic example of slicing. A varA = C(); leaves you with object of static and dynamic type of A. As a result, C* varC = static_cast<C*>(&varA); exhibits undefined behavior.

Answer 2

You can store a complete derived class in a base-class pointer, however:
int main() {
A* varA = new C();
C* varC = static_cast<C*>(varA);
varC->f();
cout << "varC->a is " << varC->a << endl;
cout << "varC->b is " << varC->b << endl;
cout << "varC->c is " << varC->c << endl;
} // oops, forgot to delete varA/varC, memory leak!