Explore Exams
Login Signup
Sign in Open App
Software Development Exam  >  Software Development Test  >  DSA in C++  >  Test: Object Oriented Programming - 2 - Software Development MCQ

Object Oriented Programming - 2 - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Object Oriented Programming - 2 (15 Questions)

You can prepare effectively for Software Development DSA in C++ with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Object Oriented Programming - 2". These 15 questions have been designed by the experts with the latest curriculum of Software Development 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 15

Sign up on EduRev for free to attempt this test and track your preparation progress.

Test: Object Oriented Programming - 2 - Question 1
Save

Which of the following is not a principle of Object-Oriented Programming (OOP)?

Detailed Solution: Question 1

Normalization is not a principle of Object-Oriented Programming (OOP). The correct principles are Inheritance, Encapsulation, and Abstraction.

Test: Object Oriented Programming - 2 - Question 2
Save

Which keyword is used to create an object in C++?

Detailed Solution: Question 2

The "new" keyword is used to dynamically allocate memory and create an object in C++.

Clear all your doubts with EduRev
Test: Object Oriented Programming - 2 - Question 3
Save

In C++, a class is a:

Detailed Solution: Question 3

A class in C++ serves as a blueprint for creating objects. It defines the properties (member variables) and behaviors (member functions) that an object of that class can have.

Test: Object Oriented Programming - 2 - Question 4
Save

What is the purpose of a constructor in C++?

Detailed Solution: Question 4

Constructors are special member functions in a class that are used to initialize the member variables of an object when it is created.

Test: Object Oriented Programming - 2 - Question 5
Save

Which access specifier allows the members of a class to be accessible from any other class?

Detailed Solution: Question 5

The "public" access specifier allows the members of a class to be accessible from any other class.

Test: Object Oriented Programming - 2 - Question 6
Save

What will be the output of the following code?
#include <iostream>

class MyClass {
public:
    int x;

    void display() {
        std::cout << "x = " << x << std::endl;
    }
};

int main() {
    MyClass obj;
    obj.x = 5;
    obj.display();

    return 0;
}

Detailed Solution: Question 6

The code creates an object of the class "MyClass" and assigns a value of 5 to the member variable "x". Then, the "display" function is called, which prints the value of "x" as 5.

Test: Object Oriented Programming - 2 - Question 7
Save

What will be the output of the following code?

#include <iostream>

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

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

int main() {
    Derived d;
    Base* ptr = &d;
    ptr->show();

    return 0;
}

Detailed Solution: Question 7

The code creates an object of the class "Derived" and assigns its address to a pointer of type "Base". When the "show" function is called using the pointer, it invokes the function from the derived class, printing "Derived class".

Test: Object Oriented Programming - 2 - Question 8
Save

What will be the output of the following code?
#include <iostream>

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

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

int main() {
    Derived d;
    Base* ptr = &d;
    ptr->show();

    return 0;
}

Detailed Solution: Question 8

The code is similar to the previous question, but the "show" function in the base class is declared as virtual. This enables dynamic polymorphism, so when the function is called through the base class pointer, it invokes the function from the derived class, printing "Derived class".

Test: Object Oriented Programming - 2 - Question 9
Save

What will be the output of the following code?
#include <iostream>

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

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

int main() {
    Derived d;
    Derived* ptr = &d;
    Base* bptr = ptr;
    bptr->show();

    return 0;
}

Detailed Solution: Question 9

The code creates an object of the class "Derived" and assigns its address to a pointer of type "Derived". Then, the pointer is implicitly cast to a base class pointer ("Base* bptr = ptr;"). When the "show" function is called through the base class pointer, it invokes the function from the derived class, printing "Derived class".

Test: Object Oriented Programming - 2 - Question 10
Save

What will be the output of the following code?

#include <iostream>

class Base {
public:
    int x;

    Base() {
        x = 0;
    }

    virtual void display() {
        std::cout << "x = " << x << std::endl;
    }
};

class Derived : public Base {
public:
    Derived() {
        x = 5;
    }

    void display() {
        std::cout << "x = " << x << std::endl;
    }
};

int main() {
    Base* bptr = new Derived();
    bptr->display();

    delete bptr;
    return 0;
}

Detailed Solution: Question 10

The code creates a dynamic object of the derived class "Derived" using a base class pointer ("Base* bptr = new Derived();"). When the "display" function is called through the base class pointer, it invokes the overridden function in the derived class, printing "x = 5". The object is properly deallocated using "delete" at the end.

Test: Object Oriented Programming - 2 - Question 11
Save

Which type of inheritance is illustrated in the following code?
class A {
public:
    void foo() {
        std::cout << "A" << std::endl;
    }
};

class B : public A {
public:
    void foo() {
        std::cout << "B" << std::endl;
    }
};

int main() {
    B b;
    b.foo();

    return 0;
}

Detailed Solution: Question 11

The code demonstrates single inheritance where class B inherits from class A. The derived class B has a function named "foo" which overrides the function with the same name in the base class A.

Test: Object Oriented Programming - 2 - Question 12
Save

Which statement is true about function overriding in C++?

Detailed Solution: Question 12

Function overriding allows a derived class to provide a different implementation of a base class function. It is achieved by declaring a function with the same name and signature in the derived class.

Test: Object Oriented Programming - 2 - Question 13
Save

What is the output of the following code?
#include <iostream>

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

class B : public A {
public:
    void foo() {
        std::cout << "B" << std::endl;
    }
};

int main() {
    A* ptr = new B();
    B* bptr = dynamic_cast<B*>(ptr);
    bptr->foo();

    return 0;
}

Detailed Solution: Question 13

The code creates a pointer of type "A*" and assigns the address of a dynamically allocated object of the derived class "B" to it. The pointer is then cast to a "B*" using dynamic_cast. When the "foo" function is called using the derived class pointer, it invokes the function from the derived class, printing "B".

Test: Object Oriented Programming - 2 - Question 14
Save

Which statement is true about pure virtual functions in C++?

Detailed Solution: Question 14

A class containing a pure virtual function is called an abstract class. An abstract class cannot be instantiated, meaning objects of the abstract class cannot be created. However, a derived class can inherit from the abstract class and provide implementations for the pure virtual functions.

Test: Object Oriented Programming - 2 - Question 15
Save

Which concept is illustrated in the following code?
class Base {
public:
    virtual void foo() = 0;
};

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

int main() {
    Base* ptr = new Derived();
    ptr->foo();

    return 0;
}

Detailed Solution: Question 15

The code demonstrates polymorphism by using a base class pointer to invoke a function that is overridden in the derived class. This allows the program to exhibit different behaviors based on the actual object type at runtime. In this case, the "foo" function is invoked through the base class pointer, and it prints "Derived" because the object is of the derived class type.

152 videos|118 docs|24 tests
Join Course
Information about Test: Object Oriented Programming - 2 Page
In this test you can find the Exam questions for Test: Object Oriented Programming - 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Object Oriented Programming - 2, EduRev gives you an ample number of Online tests for practice
152 videos|118 docs|24 tests
Join Course
Download as PDF

About this Software Development Practice Test

This test contains 15 MCQs on Object Oriented Programming - 2 with detailed solutions for each question. It is part of the DSA in C++ course on EduRev, designed to align with the current Software Development syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

Explore more Software Development Study Resources

If you found this Object Oriented Programming - 2 MCQ Test helpful, you can further enhance your Software Development preparation with in-depth courses on EduRev, for all the subjects and topics available in DSA in C++ course. The course offers:
  • Video Lectures: covering all the topics of DSA in C++
  • Notes for Revision: offering revision notes, important questions, summaries, flashcards, cheatsheets & more for Software Development preparation
  • MCQ Test: variety of questions as per the pattern of Software Development exam
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily