Base class pointers in C++ - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a base class pointer.

C++

Base* [1];

Drag options to blanks, or click blank then click option'

AbasePtr

Bobj

Cpointer

Dptr

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is not a pointer.

Forgetting the asterisk (*) to declare a pointer.

The pointer to the base class is declared as 'basePtr' here, which is a common naming convention.

2fill in blank

medium

Complete the code to assign the address of a derived class object to a base class pointer.

C++

Derived d;
Base* ptr = [1];

Drag options to blanks, or click blank then click option'

A&d

Bd

Cptr

D*d

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning the object itself instead of its address.

Using the pointer variable name instead of the object.

We assign the address of the derived object 'd' to the base class pointer using '&d'.

3fill in blank

hard

Fix the error in the code to correctly call the base class function using a base class pointer.

C++

Base* ptr = new Derived();
ptr->[1]();

Drag options to blanks, or click blank then click option'

AderivedFunction

BbaseFunction

CDerivedFunction

DBaseFunction

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to call a derived class specific function through a base class pointer.

Using incorrect capitalization for function names.

The base class pointer can only call functions declared in the base class, so 'baseFunction' is correct.

4fill in blank

hard

Fill both blanks to create a base class pointer that points to a derived class object and calls a virtual function.

C++

Base* [1] = new Derived();
[2]->virtualFunction();

Drag options to blanks, or click blank then click option'

Aptr

BbasePtr

CderivedPtr

DobjPtr

Attempts:

3 left

💡 Hint

Common Mistakes

Using different pointer names in declaration and call.

Calling non-virtual functions expecting polymorphic behavior.

The pointer 'basePtr' is used to point to the derived object and call the virtual function correctly.

5fill in blank

hard

Fill all three blanks to correctly delete a derived class object through a base class pointer and avoid memory leaks.

C++

Base* [1] = new Derived();
// ... use the object
delete [2];
[3] = nullptr;

Drag options to blanks, or click blank then click option'

Aptr

BbasePtr

CderivedPtr

DobjPtr

Attempts:

3 left

💡 Hint

Common Mistakes

Deleting using a different pointer than the one used to allocate.

Not setting the pointer to nullptr after deletion.

Deleting the object through the base class pointer 'basePtr' and setting it to nullptr prevents memory leaks.

Practice

1. What does a base class pointer in C++ allow you to do?

easy

A. Only point to base class objects

B. Point to objects of derived classes

C. Only point to primitive data types

D. Prevent any pointer arithmetic

Solution

1. Step 1: Understand base and derived classes

   A base class is a general class, and derived classes extend it with more features.

2. Step 2: Role of base class pointers

   A base class pointer can point to any object of the base class or its derived classes, enabling polymorphism.

3. Final Answer:

   Point to objects of derived classes -> Option B

4. Quick Check:

   Base pointer can point to derived objects [OK]

Hint: Base pointers can point to derived objects [OK]

Common Mistakes:

• Thinking base pointers only point to base objects
• Confusing pointers with references
• Assuming base pointers can't access derived objects

2. Which of the following is the correct syntax to declare a base class pointer named ptr for a class Base?

easy

A. Base *ptr;

B. ptr Base*;

C. Base ptr*;

D. *Base ptr;

Solution

1. Step 1: Recall pointer declaration syntax

   In C++, a pointer to a type is declared as Type *pointerName;.

2. Step 2: Apply to Base class

   For class Base, the pointer declaration is Base *ptr;.

3. Final Answer:

   Base *ptr; -> Option A

4. Quick Check:

   Pointer syntax = Type *name [OK]

Hint: Pointer syntax is Type *name; [OK]

Common Mistakes:

• Placing * after variable name incorrectly
• Swapping type and variable name
• Using invalid syntax like *Base ptr

3. Consider the code:

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

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

What is the output when this code runs?

medium

A. Derived

B. Base

C. Compilation error

D. Runtime error

Solution

1. Step 1: Understand virtual function behavior

   The show() function is virtual in Base, so calls use the actual object's method.

2. Step 2: Pointer points to Derived object

   The pointer ptr points to a Derived object, so Derived::show() is called.

3. Final Answer:

   Derived -> Option A

4. Quick Check:

   Virtual function calls Derived method [OK]

Hint: Virtual functions call derived version via base pointer [OK]

Common Mistakes:

• Ignoring virtual keyword and expecting Base output
• Confusing pointer type with object type
• Assuming compilation or runtime error

4. What is wrong with this code snippet?

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

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

medium

A. delete ptr causes memory leak

B. Pointer ptr cannot point to Derived object

C. Derived class must not redefine display()

D. display() should be declared virtual in Base for polymorphism

Solution

1. Step 1: Check function overriding and polymorphism

   Without virtual, base class pointer calls base version even if pointing to derived object.

2. Step 2: Identify missing virtual keyword

   To enable polymorphism, display() in Base must be declared virtual.

3. Final Answer:

   display() should be declared virtual in Base for polymorphism -> Option D

4. Quick Check:

   Missing virtual causes base method call [OK]

Hint: Declare base methods virtual for polymorphism [OK]

Common Mistakes:

• Assuming pointer can't point to derived
• Thinking delete causes leak here
• Believing derived can't redefine base methods

5. You have a base class Animal with a virtual function sound(). Two derived classes Dog and Cat override sound(). You want to store multiple animals in an array and call sound() on each. Which approach correctly uses base class pointers to achieve this?

hard

A. Create an array of Animal objects and assign Dog and Cat objects directly

B. Create an array of Dog* and cast Cat* to Dog*

C. Create an array of Animal*, store new Dog() and new Cat(), then loop and call sound()

D. Use an array of void* pointers and cast to Animal* before calling sound()

Solution

1. Step 1: Understand polymorphic storage

   To store different derived objects, use an array of base class pointers (Animal*).

2. Step 2: Store derived objects via pointers

   Store new Dog() and new Cat() in the Animal* array, then call sound() polymorphically.

3. Final Answer:

   Create an array of Animal*, store new Dog() and new Cat(), then loop and call sound() -> Option C

4. Quick Check:

   Base pointer array + virtual functions = polymorphism [OK]

Hint: Use base class pointer array for polymorphic objects [OK]

Common Mistakes:

• Using object array causes slicing
• Casting unrelated pointers unsafely
• Using void* loses type safety and polymorphism