Resolution of virtual function with default parameters [duplicate]

Answer 1

Your code is actually seen by the compiler like this:
(The display() method is not actually there, but the resolving works in similar way)

class A
{
public:
    virtual void display(int i) { cout<< "Base::" << i << endl; }
    void display() { display(5); }
};

class B : public A
{
public:
    void display(int i) override { cout<< "Derived::" << i << endl; }
    void display() { display(9); }
};

Now you should understand what happens. You are calling the non-virtual display() which calls the virtual function. In more strict words: the default argument is resolved just like if the no-arg non-virtual method was there - by the type of the variable (not by the actual type of the object), but the code gets executed according to real object type, because it is virtual method:

int main()
{
    A * a = new B(); // type of a is A*   real type is B
    a->display();    // calls A::display() which calls B::display(5)

    A* aa = new A(); // type of aa is A*  real type is A
    aa->display();   // calls A::display() which calls A::display(5)

    B* bb = new B(); // type of bb is B*  real type is B
    bb->display();   // calls B::display() which calls B::display(9)
}