C++ Shadowing Base Class Member Function

As we know, inheritance is a feature of OOP that allows us to create derived classes from a base class. The derived classes inherit features of the base class.

Suppose we define the same function in both the base class and the derived class. Now, when we call this function using the object of the derived class, the function of the derived class executes.

Here, the member function in the derived class shadows the member function in the base class. This is called shadowing the base class member function

Example 1: C++ Shadowing Base Class Member Function

// C++ program to demonstrate shadowing base class member function

#include <iostream>
using namespace std;

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

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

int main() {
    Derived derived1;
    derived1.print();
    return 0;
}

Output

Derived Function

Here, the same function print() is defined in both Base and Derived classes.

So, when we call print() from the Derived object derived1, the print() from Derived is executed by shadowing the function in Base.

As we can see, the function was shadowed because we called the function from an object of the Derived class.

Had we called the print() function from an object of the Base class, the function would not have been shadowed.

// Call function of Base class
Base base1;
base1.print(); // Output: Base Function

Access Shadowed Function in C++

To access the shadowed function of the base class, we use the scope resolution operator ::.

We can also access the shadowed function by using a pointer of the base class to point to an object of the derived class and then calling the function from that pointer.

Example 2: C++ Access Shadowed Function From the Base Class

// C++ program to access shadowed function
// in main() using the scope resolution operator ::

#include <iostream>
using namespace std;

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

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

int main() {
    Derived derived1, derived2;
    derived1.print();

    // access print() function of the Base class
    derived2.Base::print();

    return 0;
}

Output

Derived Function
Base Function

Here, the statement

derived2.Base::print();

accesses the print() function of the Base class.

Example 3: C++ Call Shadowed Function From Derived Class

// C++ program to call the shadowed function
// from a member function of the derived class

#include <iostream>
using namespace std;

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

class Derived : public Base {
   public:
    void print() {
        cout << "Derived Function" << endl;

        // call overridden function
        Base::print();
    }
};

int main() {
    Derived derived1;
    derived1.print();
    return 0;
}

Output

Derived Function
Base Function

In this program, we have called the base class member function inside the Derived class itself.

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

Notice the code Base::print();, which calls the member function print() from the Base class inside the Derived class.

Example 4: C++ Call Shadowed Function Using Pointer

// C++ program to access shadowed function using pointer
// of Base type that points to an object of Derived class

#include <iostream>
using namespace std;

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

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

int main() {
    Derived derived1;

    // pointer of Base type that points to derived1
    Base* ptr = &derived1;

    // call function of Base class using ptr
    ptr->print();

    return 0;
}

Output

Base Function

In this program, we have created a pointer of Base type named ptr. This pointer points to the Derived object derived1.

// pointer of Base type that points to derived1
Base* ptr = &derived1;

When we call the print() function using ptr, it calls the member function from Base.

// call function of Base class using ptr
ptr->print();

This is because even though ptr points to a Derived object, it is actually of Base type. So, it calls the member function of Base.

In order to override the Base function instead of accessing it, we need to use virtual functions in the Base class.