Access control in inheritance in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of access control with public inheritance

What is the output of this C++ code snippet?

C++

#include <iostream>
class Base {
public:
    int x = 10;
protected:
    int y = 20;
private:
    int z = 30;
};
class Derived : public Base {
public:
    void print() {
        std::cout << x << ' ' << y << '\n';
        // std::cout << z << '\n'; // z is private in Base
    }
};
int main() {
    Derived d;
    d.print();
    std::cout << d.x << '\n';
    // std::cout << d.y << '\n'; // y is protected
    return 0;
}

A 10 20\n20\n

B 10 20\n10\n

C 10 30\n10\n

D Compilation error due to access of y

💡 Hint

Remember that public inheritance keeps public members public and protected members protected in the derived class.

In public inheritance, public members remain public and protected members remain protected in the derived class. The private members of the base class are not accessible in the derived class. So, x and y can be accessed inside Derived, but z cannot. Outside the class, only public members are accessible, so d.x works but d.y does not.

❓ Predict Output

intermediate

Access control with protected inheritance

What will be the output or error of this C++ code?

C++

#include <iostream>
class Base {
public:
    int a = 5;
protected:
    int b = 10;
};
class Derived : protected Base {
public:
    void show() {
        std::cout << a << ' ' << b << '\n';
    }
};
int main() {
    Derived d;
    d.show();
    // std::cout << d.a << '\n'; // Is this allowed?
    return 0;
}

A 5 10\nCompilation error on accessing d.a

B 5 10\n5\n

C Compilation error inside show() due to access of a

D Compilation error inside show() due to access of b

💡 Hint

Protected inheritance changes the access level of base class public members to protected in the derived class.

With protected inheritance, the public and protected members of the base class become protected in the derived class. So inside Derived, both a and b are accessible. But outside, d.a is not accessible because it is now protected in Derived. Hence, the code inside show() runs fine, but accessing d.a in main() causes a compilation error.

❓ Predict Output

advanced

Private inheritance and member accessibility

What will happen when compiling and running this code?

C++

#include <iostream>
class Base {
public:
    int val = 100;
};
class Derived : private Base {
public:
    void print() {
        std::cout << val << '\n';
    }
};
int main() {
    Derived d;
    d.print();
    // std::cout << d.val << '\n'; // Is this allowed?
    return 0;
}

A Compilation error: val is private in Base

B Compilation error inside print() due to access of val

C 100\n100\n

D 100\nCompilation error on accessing d.val

💡 Hint

Private inheritance makes all base class public and protected members private in the derived class.

With private inheritance, all public and protected members of the base class become private in the derived class. So inside Derived, val is accessible. But outside, d.val is not accessible because it is private in Derived. Therefore, print() works and prints 100, but accessing d.val in main() causes a compilation error.

🧠 Conceptual

advanced

Effect of inheritance type on base class member accessibility

Given a base class with public, protected, and private members, which inheritance type will make the base class public members inaccessible from outside the derived class?

A Private inheritance

B Protected inheritance

C Public inheritance

D Virtual inheritance

💡 Hint

Think about which inheritance type changes public members to private in the derived class.

In private inheritance, all public and protected members of the base class become private in the derived class. This means they are not accessible from outside the derived class. Public and protected inheritance keep public members accessible outside the derived class. Virtual inheritance affects the way base classes are shared but does not change access levels.

🔧 Debug

expert

Why does this code fail to compile?

This code fails to compile. What is the cause of the error?

C++

#include <iostream>
class Base {
protected:
    int data = 42;
};
class Derived : public Base {
public:
    void show() {
        std::cout << data << '\n';
    }
};
int main() {
    Derived d;
    d.show();
    std::cout << d.data << '\n';
    return 0;
}

A No error, code compiles and prints 42 twice

B Error because 'data' is private in Base

C Error because 'data' is protected and cannot be accessed outside the class or derived classes

D Error because 'show' is not declared const

💡 Hint

Check the access level of 'data' and where it is accessed.

The member data is protected in Base. This means it can be accessed inside Base and its derived classes, but not from outside. The call d.show() works because show() is a member of Derived which can access data. However, std::cout << d.data in main() tries to access data from outside, which is not allowed and causes a compilation error.