Обеспечение более строгого доступа к производному классу В C++

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Hi everyone i have one question is my mind

This program worked
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#include<iostream>
using namespace std;
//declaring base class
class Base {
public:
    virtual int fun(int i) { cout << "Base::fun(int i) called"; }
};
//Declaring Derived class
class Derived: public Base {
private:
    int fun(int x)   { cout << "Derived::fun(int x) called"; }
};
int main()
{
    Base *ptr = new Derived;
    ptr->fun(10);
    return 0;
}

//It worked and gave output as :
Derived::fun(int x) called

But the below program didn't worked, don't know why
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#include<iostream>
using namespace std;
//Declaring base class
class Base {
private:
    virtual int fun(int i) { cout << "Base::fun(int i) called"; }
};
//Declaring Derived class
class Derived: public Base {
public:
    int fun(int x)   { cout << "Derived::fun(int x) called"; }
};
int main()
{
    Base *ptr = new Derived;
    ptr->fun(10);
    return 0;
}

Что я уже пробовал:

Я изучил виртуальные функции, но запутался в работе двух вышеперечисленных программ

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