Question 1

If a base class has a virtual function named print, and a pointer variable of that class is pointing to a derived object, then the code ptr->print ); calls

Accepted Answer

The virtual function keyword ensures that the derived class implementation of the function is called when a pointer to the derived class is being used. Therefore, the code ptr->print(); will call the print function of the derived class.

Question 2

If you have a copy constructor in the base class, but do not have a copy constructor for the derived class, then

Accepted Answer

If you have a copy constructor in the base class but not in the derived class, then the compiler will automatically create a copy constructor for the derived class which will call the copy constructor of the base class. This behavior is known as "derived class copy constructor synthesis." Therefore, option B is the correct answer.

Question 3

If a base class has a non-virtual member function named print, and a pointer variable of that class is pointing to a derived object, then the code ptr->print ); calls

Accepted Answer

Since the member function is non-virtual, the function call is resolved at compile-time based on the type of the pointer variable, which is of the base class type. Therefore, the base class print function is called, regardless of whether the pointer variable points to a base object or a derived object.

Question 4

If a derived class Class2) has redefined a function from the base class Class 1), how can that derived function call the base class function if the function declaration is as follows?
Void print );

Accepted Answer

To call the base class function, we can use the function name prefixed by the base class name followed by the scope resolution operator "::". Therefore, the correct syntax would be Class1::print(). The option A is incorrect as it suggests that we should publicly inherit the base class function, which is not required to call the base class function. Option C is also incorrect as it only specifies the derived class function name without indicating that it needs to call the base class function. Option D is incorrect as only option B is the correct syntax for calling the base class function.

Question 5

Given the following classes and code, what is the output of the last statement shown? Class Pet { Public: Virtual void print); String name; Private: }; Class Dog: public Pet { Public: Void print); String breed; }; Void Pet::print) { Cout << "My name is " << name; } Void Dog::print) { Pet::print); Cout << ", and my breed is a "<< breed << endl; } Pet* pPtr; Dog* dPtr=new Dog; DPtr->name= "Rover"; DPtr->breed="Weiner"; PPtr= dPtr; PPtr->print);

Accepted Answer

The code creates a Dog object with name "Rover" and breed "Weiner". The print() method is called on the Pet pointer pPtr, which has been assigned to the Dog object dPtr. Since pPtr is of type Pet pointer, it calls the Pet class's print() method, which outputs "My name is Rover". Then the Dog class's print() method is called, which outputs ", and my breed is a Weiner". Therefore, the output is "My name is Rover, and my breed is a Weiner".

Question 6

Given the following classes and code, what is the output of the last statement shown? Class Pet { Public: Virtual void print); String name; Private: }; Class Dog: public Pet { Public: Void print); String breed; }; Void Pet::print) { Cout << "My name is " << name; } Void Dog::print) { Pet::print); Cout << ", and my breed is a "<< breed << endl; } Pet pPtr; Dog dPtr; DPtr->name= "Rover"; DPtr->breed="Weiner"; PPtr= dPtr; PPtr->print);

Accepted Answer

The `print` function of the `Pet` class is called because `pPtr` is of type `Pet`, and the `print` function is not virtual. Therefore, it outputs "My name is Rover".

Question 7

Given a class A that derives from a class B that derives from a class C, when an object of class A goes out of scope, in which order are the destructors called?

Accepted Answer

Destructors are called in the reverse order of constructors. Since A's constructor is called first, A's destructor will be called last, followed by B's destructor and finally C's destructor. Therefore, the order of destruction will be C, B, then A.

Question 8

Which of the following would correctly call the base class BaseClass) assignment operator from the derived class DerivedClass) assignment operator?
DerivedClass& DerivedClass::operator =const DerivedClass& rightSide)
{
//what goes here?
}

Accepted Answer

To call the base class assignment operator from the derived class assignment operator, we use the scope resolution operator (::) to specify the base class and then call the base class assignment operator passing in the right-hand side of the assignment. Thus, BaseClass::operator = (rightSide) is the correct choice.

Question 9

Which of the following should be virtual if a base class uses dynamic memory allocation?

Accepted Answer

If a base class uses dynamic memory allocation, then its derived classes also inherit this behavior. Therefore, the destructor should be virtual to ensure that the derived class's destructor gets called when an object is destroyed. This helps to prevent memory leaks and ensure proper cleanup of dynamically allocated memory. The other options are not necessary to be virtual in this scenario.

Question 10

In order to tell the compiler to wait to decide which version of a function to use, you must precede the function declaration in the base class with the keyword

Accepted Answer

The keyword "virtual" is used to indicate that a function may be overridden by a derived class, and that the decision on which version of the function to use should be made at runtime based on the actual type of the object being referred to. This allows for polymorphism and dynamic dispatch, which are important features of object-oriented programming.

Question 11

You should make a function a virtual function if

Accepted Answer

Virtual functions are used when derived classes need to override the function with their own implementation. Therefore, if every derived class needs to redefine the function, it is appropriate to make it virtual.

Question 12

Polymorphism refers to

Accepted Answer

Polymorphism allows a function or method to be able to represent multiple types while still maintaining its functionality. This is often achieved through function overloading or function overriding, but the key aspect is that a single function name may have multiple implementations or behaviors based on the context in which it is used. Option B refers specifically to function overriding, while option C refers specifically to function overloading, both of which are examples of how polymorphism can be achieved. However, neither of these options captures the full meaning of polymorphism.

Question 13

Given the following simplified classes,
Class Pet
{
Public:
Virtual void print);
String name;
Private:
};
Class Dog: public Pet
{
Public:
Void print);
String breed;
};
Dog vDog;
Pet vPet;
VDog.name="rover";
VDog.breed = "Collie";
Which of the following statements are not legal?

Accepted Answer

Option D is not legal because the base class Pet does not have a member variable called breed. Therefore, it would not be possible to access the breed member variable of the derived class Dog through the base class Pet object. Options A, B, and C are legal because the Pet class has a member variable called name, which is inherited by the derived class Dog, and can be accessed through a base-class reference to a derived-class object. Option A will print "rover" because it is accessing the name member variable of the Dog object through a base-class reference. Option B will not compile because Dog's breed member is being accessed through vDog, which is a base-class reference. Option C will print "rover" because vPet is assigned the value of VDog and print makes a virtual call to Dog's print method.

Quiz 15: Inheritance

If a base class has a virtual function named print, and a pointer variable of that class is pointing to a derived object, then the code ptr->print ) ; calls

If you have a copy constructor in the base class, but do not have a copy constructor for the derived class, then

If a base class has a non-virtual member function named print, and a pointer variable of that class is pointing to a derived object, then the code ptr->print ) ; calls

If a derived class Class2) has redefined a function from the base class Class 1) , how can that derived function call the base class function if the function declaration is as follows? Void print ) ;

Given a class A that derives from a class B that derives from a class C, when an object of class A goes out of scope, in which order are the destructors called?

Which of the following would correctly call the base class BaseClass) assignment operator from the derived class DerivedClass) assignment operator? DerivedClass& DerivedClass::operator =const DerivedClass& rightSide) { //what goes here? }

Which of the following should be virtual if a base class uses dynamic memory allocation?

In order to tell the compiler to wait to decide which version of a function to use, you must precede the function declaration in the base class with the keyword

You should make a function a virtual function if

Polymorphism refers to

Given the following simplified classes, Class Pet { Public: Virtual void print) ; String name; Private: }; Class Dog: public Pet { Public: Void print) ; String breed; }; Dog vDog; Pet vPet; VDog.name="rover"; VDog.breed = "Collie"; Which of the following statements are not legal?