On name lookup in C++ inheritance

Examples are as follows:

#include<iostream>
#include<string>
using namespace std;
class Base {
 public:
 void func() {
  cout << "func() in Base." << endl;
 }
 void func(int a) {
  cout << "func(int a) in Base." << endl;
 }
 void func(string s) {
  cout << "func(string s) in Base." << endl;
 }
};


class Derived : public Base { 
public:
 //using Base::func;
 void print() {
  cout << "func() in Derived." << endl;
 }
};


int main() {
 Derived d;
 d.Base::func();// Specify the base class version 
 d.func();
 d.func(12);//error, Can be added in a derived class using Base::print;
 d.func("abc");//error Can be added to a derived class using Base::print;
 system("pause");
 return 0;
}
//1 The static types of objects, references, and Pointers determine which members of the object are visible. 
// The derived class scope is nested within the base class scope 
// Derived class members mask base class members with the same name 
// If the derived class wants to use an overloaded version of a base class by its own type , The derived class must either override all overloaded versions , either 1 Not one of them. 
// using using The declaration adds all overloaded versions of the base class to the derived class scope 
// The name lookup precedes type checking, and if the name of the function is found in the derived class, the lookup does not proceed up, followed by type checking