Derived class hiding base class same name method - c++

Question

I was reading this article and it states
This calls Derived::f( complex ). Why? Well, remember that
Derived doesn't declare "using Base:f;", and so clearly Base::f( int )
and Base::f( double ) can't be called.
I decided to try this out and used this code
class Base {
public:
virtual void f( int ) {
cout << "Base::f(int)" << endl;
}
virtual void f( double ) {
cout << "Base::f(double)" << endl;
}
virtual void g( int i = 10 ) {
cout << i << endl;
}
};
class Derived: public Base {
using Base::f;
public:
void f( complex<double> ) {
cout << "Derived::f(complex)" << endl;
}
void g( int i = 20 ) {
cout << "Derived::g() " << i << endl;
}
};
int main() {
Derived d;
d.f(1.0);
}
I get the error
main.cpp: In function 'int main()':
main.cpp:43:16: error: 'virtual void Base::f(double)' is inaccessible within this context
43 | d.f(1.0);
My question is how do I use using Base::f; and how do I fix this issue ?

Answer 1

Simple. You should write the 'using' declaration under public section of class as below. If you put the declaration in private section, Base functions are available to Derived class but they will become private members of Derived class. That is why your compiler is throwing 'inaccessible' error.
#include <iostream>
#include <complex>
using namespace std;
class Base {
public:
virtual void f( int ) {
cout << "Base::f(int)" << endl;
}
virtual void f( double ) {
cout << "Base::f(double)" << endl;
}
virtual void g( int i = 10 ) {
cout << i << endl;
}
};
class Derived: public Base {
public:
using Base::f;
void f( complex<double> ) {
cout << "Derived::f(complex)" << endl;
}
void g( int i = 20 ) {
cout << "Derived::g() " << i << endl;
}
};
int main() {
Derived d;
d.f(1.0);
}