Function overriding in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of overridden function call via base pointer

What is the output of this C++ code when calling the function through a base class pointer?

C++

#include <iostream>
using namespace std;

class Base {
public:
    virtual void show() { cout << "Base show" << endl; }
};

class Derived : public Base {
public:
    void show() override { cout << "Derived show" << endl; }
};

int main() {
    Base* ptr = new Derived();
    ptr->show();
    delete ptr;
    return 0;
}

A Base show

B Derived show

C Runtime error

D Compilation error

💡 Hint

Remember that virtual functions allow the derived class version to be called through a base pointer.

Because show() is declared virtual in Base and overridden in Derived, calling show() through a Base pointer to a Derived object calls Derived's version.

❓ Predict Output

intermediate

Output when overriding function is not virtual

What will this C++ program print when calling the function through a base class pointer?

C++

#include <iostream>
using namespace std;

class Base {
public:
    void display() { cout << "Base display" << endl; }
};

class Derived : public Base {
public:
    void display() { cout << "Derived display" << endl; }
};

int main() {
    Base* ptr = new Derived();
    ptr->display();
    delete ptr;
    return 0;
}

A Runtime error

B Compilation error

C Derived display

D Base display

💡 Hint

Check if the function is declared virtual in the base class.

Since display() is not virtual, the function called depends on the pointer type, not the object type. The base class version is called.

🔧 Debug

advanced

Identify the error in function overriding

What error does this C++ code produce when compiled?

C++

#include <iostream>
using namespace std;

class Base {
public:
    virtual void func(int x) { cout << "Base: " << x << endl; }
};

class Derived : public Base {
public:
    void func(double x) override { cout << "Derived: " << x << endl; }
};

int main() {
    Derived d;
    d.func(5);
    return 0;
}

A Compilation error: 'func' marked override but does not override

B Runtime error: ambiguous call

C Output: Derived: 5

D Output: Base: 5

💡 Hint

Check if the function signature exactly matches the base class virtual function.

The Derived class's func(double) does not match Base's func(int), so override keyword causes a compilation error.

🧠 Conceptual

advanced

Effect of 'final' on function overriding

What happens if a virtual function in a base class is marked with 'final' and a derived class tries to override it?

A Derived class function overrides successfully

B Runtime error when calling the function

C Compilation error: overriding a final function is not allowed

D The base class function is hidden but still called

💡 Hint

The 'final' keyword prevents further overriding.

Marking a virtual function as final in the base class forbids derived classes from overriding it, causing a compilation error if attempted.

❓ Predict Output

expert

Output of multiple inheritance with overridden functions

What is the output of this C++ program?

C++

#include <iostream>
using namespace std;

class A {
public:
    virtual void show() { cout << "A" << endl; }
};

class B : public A {
public:
    void show() override { cout << "B" << endl; }
};

class C : public A {
public:
    void show() override { cout << "C" << endl; }
};

class D : public B, public C {
public:
    void show() override { cout << "D" << endl; }
};

int main() {
    D obj;
    obj.show();
    B* bptr = &obj;
    bptr->show();
    C* cptr = &obj;
    cptr->show();
    return 0;
}

A

D
B
C

B

D
D
D

C Compilation error due to ambiguity

D

D
A
A

💡 Hint

Each base pointer calls its own overridden version; D overrides show() for itself.

Class D overrides show(), so obj.show() prints D. bptr points to B part of obj, calling B::show(). cptr points to C part, calling C::show().