Interface-like behavior in C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is an interface in C++?

An interface in C++ is a class that has only pure virtual functions and no data members. It defines a contract that other classes can implement.

beginner

How do you declare a pure virtual function in C++?

You declare a pure virtual function by assigning 0 to the function declaration, like this: virtual void func() = 0;.

beginner

Why can't you create an object of a class with pure virtual functions?

Because such a class is abstract and incomplete. It requires derived classes to provide implementations for the pure virtual functions before objects can be created.

intermediate

How does C++ achieve interface-like behavior without a dedicated interface keyword?

By using abstract classes with only pure virtual functions, C++ simulates interfaces. Classes implementing the interface inherit and override these functions.

intermediate

What is the role of a virtual destructor in an interface-like class?

A virtual destructor ensures that when deleting an object through a pointer to the interface, the derived class's destructor is called properly, avoiding resource leaks.

Which of the following declares a pure virtual function in C++?

A void foo() {}

B virtual void foo() = 0;

C virtual void foo();

D void foo() = 0;

Only 'virtual void foo() = 0;' declares a pure virtual function.

Can you instantiate an object of a class that has at least one pure virtual function?

A No, never.

B Only if the class has a constructor.

C Only if the pure virtual function has a body.

D Yes, always.

Classes with pure virtual functions are abstract and cannot be instantiated.

What keyword does C++ use to indicate a function is virtual?

A virtual

B interface

C abstract

D override

The 'virtual' keyword marks a function as virtual in C++.

Why should an interface-like class have a virtual destructor?

A To make all functions pure virtual.

B To prevent the class from being instantiated.

C To allow derived class destructors to run correctly when deleting via base pointer.

D To improve performance.

A virtual destructor ensures proper cleanup of derived objects.

Which of these best describes an interface-like class in C++?

A A class with private constructors.

B A class with no functions.

C A class with only static functions.

D A class with only pure virtual functions and no data members.

Interface-like classes have only pure virtual functions and no data members.

Explain how C++ uses abstract classes to simulate interfaces.

Think about classes with functions set to zero.

Why is it important to have a virtual destructor in an interface-like class?

Consider what happens when deleting objects via base class pointers.

Practice

1. What is the main purpose of an interface-like class in C++?

easy

A. To store data members for derived classes

B. To provide full function implementations for reuse

C. To define functions that derived classes must implement

D. To create objects directly without inheritance

Solution

1. Step 1: Understand interface concept

   Interfaces specify what functions a class must have but do not provide their code.

2. Step 2: Identify interface role in C++

   In C++, interfaces are created using pure virtual functions that derived classes must implement.

3. Final Answer:

   To define functions that derived classes must implement -> Option C

4. Quick Check:

   Interface = function declarations only [OK]

Hint: Interfaces declare required functions without code [OK]

Common Mistakes:

• Thinking interfaces provide function code
• Assuming interfaces hold data members
• Believing interfaces can be instantiated

2. Which of the following is the correct syntax to declare a pure virtual function in a C++ class?

easy

A. virtual void func() = 0;

B. virtual void func();

C. void func() override;

D. virtual void func() = 1;

Solution

1. Step 1: Recall pure virtual function syntax

   Pure virtual functions are declared with '= 0' after the function signature.

2. Step 2: Check options for correct syntax

   virtual void func() = 0; uses '= 0' correctly, others do not match pure virtual syntax.

3. Final Answer:

   virtual void func() = 0; -> Option A

4. Quick Check:

   Pure virtual = '= 0' [OK]

Hint: Pure virtual functions end with '= 0;' [OK]

Common Mistakes:

• Using '= 1' instead of '= 0'
• Omitting 'virtual' keyword
• Confusing override with pure virtual

3. Consider the following code:

class IShape {
public:
    virtual double area() const = 0;
};

class Circle : public IShape {
    double radius;
public:
    Circle(double r) : radius(r) {}
    double area() const override { return 3.14 * radius * radius; }
};

int main() {
    Circle c(2);
    std::cout << c.area();
    return 0;
}

What is the output of this program?

medium

A. 4

B. Compilation error

C. 3.14

D. 12.56

Solution

1. Step 1: Calculate area of circle with radius 2

   Area = π * r² = 3.14 * 2 * 2 = 12.56

2. Step 2: Confirm code compiles and runs correctly

   Circle implements pure virtual function area(), so no error; output is area value.

3. Final Answer:

   12.56 -> Option D

4. Quick Check:

   Area = 3.14 * 4 = 12.56 [OK]

Hint: Calculate area formula carefully [OK]

Common Mistakes:

• Forgetting to implement pure virtual function
• Using wrong radius value
• Expecting compilation error

4. What is wrong with this code snippet?

class IAnimal {
public:
    virtual void speak() = 0;
};

class Dog : public IAnimal {
public:
    void speak() { std::cout << "Woof"; }
};

int main() {
    IAnimal a;
    a.speak();
    return 0;
}

medium

A. Missing override keyword in Dog::speak()

B. Cannot create object of abstract class IAnimal

C. speak() should return a value

D. Dog class must be abstract

Solution

1. Step 1: Identify abstract class usage

   IAnimal has a pure virtual function, so it is abstract and cannot be instantiated.

2. Step 2: Check main() for object creation

   Code tries to create IAnimal a; which is illegal and causes compilation error.

3. Final Answer:

   Cannot create object of abstract class IAnimal -> Option B

4. Quick Check:

   Abstract class objects cannot be created [OK]

Hint: Abstract classes cannot be instantiated [OK]

Common Mistakes:

• Ignoring abstract class instantiation rules
• Thinking override keyword is mandatory
• Assuming pure virtual functions must return values

5. You want to create an interface-like class IPrinter with a pure virtual function print(). Then create two classes ConsolePrinter and FilePrinter that implement print(). Which of the following code snippets correctly demonstrates this?

hard

A. class IPrinter {
public:
virtual void print() = 0;
};

class ConsolePrinter : public IPrinter {
public:
void print() override { std::cout << "Console output"; }
};

class FilePrinter : public IPrinter {
public:
void print() override { /* write to file */ }
};

B. class IPrinter {
public:
void print() { }
};

class ConsolePrinter : public IPrinter {
public:
void print() { std::cout << "Console output"; }
};

C. class IPrinter {
public:
virtual void print();
};

class ConsolePrinter : public IPrinter {
};

D. class IPrinter {
public:
virtual void print() = 1;
};

Solution

1. Step 1: Define interface with pure virtual function

   IPrinter declares print() as pure virtual with '= 0', making it an interface.

2. Step 2: Implement print() in derived classes

   ConsolePrinter and FilePrinter override print() with their own code, fulfilling interface contract.

3. Final Answer:

   Code snippet D correctly defines interface and implementations -> Option A

4. Quick Check:

   Pure virtual = 0 + override implementations [OK]

Hint: Use '= 0' for pure virtual and override in derived [OK]

Common Mistakes:

• Using '= 1' instead of '= 0' for pure virtual
• Not implementing pure virtual functions in derived classes
• Defining interface functions with bodies