What are the rules for virtual function lookup?

Question

#include 
class base
{
    public:
    virtual void print (int a)
    {   
        std::cout << \"a: \" << a << \" base\\n\";
    }         


        

Answer 1

Overload resolution happens at compile time. Overrides happen at run time.

Therefore, the overload resolution of b->print(d); happens first. This selects Base::print(int) because it's the only one-argument print.

At runtime, b points to a Derived object that has no override for Base::print(int). Therefore, Base::print(int) is still called.