Base class and subclass in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of method overriding in subclass

What is the output of this Swift code?

Swift

class Animal {
    func sound() -> String {
        return "Some sound"
    }
}

class Dog: Animal {
    override func sound() -> String {
        return "Bark"
    }
}

let pet: Animal = Dog()
print(pet.sound())

A Compilation error due to missing override

B Some sound

C Bark

D Runtime error

💡 Hint

Remember that subclass methods can replace base class methods when marked with override.

The Dog class overrides the sound() method of Animal. Even though the variable is typed as Animal, the actual object is a Dog, so sound() from Dog is called, printing "Bark".

❓ Predict Output

intermediate

Value of property after subclass initialization

What will be printed by this Swift code?

Swift

class Vehicle {
    var wheels: Int
    init(wheels: Int) {
        self.wheels = wheels
    }
}

class Bicycle: Vehicle {
    init() {
        super.init(wheels: 2)
    }
}

let bike = Bicycle()
print(bike.wheels)

A 2

B 0

C Compilation error due to missing initializer

D Runtime error

💡 Hint

Check how the subclass calls the base class initializer.

The Bicycle class calls super.init(wheels: 2) to set the wheels property to 2. So printing bike.wheels outputs 2.

🔧 Debug

advanced

Identify the error in subclass initializer

What error does this Swift code produce?

Swift

class Person {
    var name: String
    init(name: String) {
        self.name = name
    }
}

class Employee: Person {
    var id: Int
    init(id: Int) {
        self.id = id
    }
}

A Error: Missing call to super.init in subclass initializer

B Error: Property 'name' not initialized before use

C No error, code compiles fine

D Error: Cannot override stored property

💡 Hint

Subclass initializers must call the base class initializer to initialize inherited properties.

The Employee initializer sets id but does not call super.init(name:). This causes a compile-time error because the inherited property name is not initialized.

❓ Predict Output

advanced

Output when calling base class method from subclass

What is the output of this Swift code?

Swift

class Printer {
    func printMessage() {
        print("Base message")
    }
}

class CustomPrinter: Printer {
    override func printMessage() {
        super.printMessage()
        print("Custom message")
    }
}

let p = CustomPrinter()
p.printMessage()

A Compilation error due to super call

B Custom message

C Base message

D

Base message
Custom message

💡 Hint

Look at how super.printMessage() is used inside the subclass method.

The CustomPrinter overrides printMessage() but calls super.printMessage() first, printing "Base message" then "Custom message" on separate lines.

🧠 Conceptual

expert

Number of stored properties in subclass instance

Given these Swift classes, how many stored properties does an instance of Manager have?

class Employee {
    var name: String
    var salary: Double
    init(name: String, salary: Double) {
        self.name = name
        self.salary = salary
    }
}

class Manager: Employee {
    var department: String
    init(name: String, salary: Double, department: String) {
        self.department = department
        super.init(name: name, salary: salary)
    }
}

A 4

B 3

C 1

D 2

💡 Hint

Count all stored properties declared in base and subclass.

An instance of Manager has name and salary from Employee, plus department from Manager, totaling 3 stored properties.

Practice

1. What is the main purpose of a base class in Swift?

easy

A. To prevent other classes from inheriting

B. To provide a general template for other classes to build upon

C. To store data only without any functions

D. To create instances that cannot be used

Solution

1. Step 1: Understand the role of a base class

   A base class defines common properties and methods that other classes can reuse.

2. Step 2: Compare options with this role

   To provide a general template for other classes to build upon correctly states the base class is a general template. Other options describe incorrect or unrelated roles.

3. Final Answer:

   To provide a general template for other classes to build upon -> Option B

4. Quick Check:

   Base class = general template [OK]

Hint: Base class = general template for subclasses [OK]

Common Mistakes:

• Thinking base class cannot be inherited
• Confusing base class with instances
• Assuming base class only stores data

2. Which of the following is the correct way to declare a subclass named Dog inheriting from a base class Animal in Swift?

easy

A. class Dog inherits Animal { }

B. class Dog extends Animal { }

C. class Dog: Animal { }

D. class Dog < Animal { }

Solution

1. Step 1: Recall Swift subclass syntax

   In Swift, a subclass is declared using a colon (:) after the subclass name followed by the base class name.

2. Step 2: Match options to syntax

   class Dog: Animal { } uses class Dog: Animal { }, which is correct. Other options use incorrect keywords or symbols.

3. Final Answer:

   class Dog: Animal { } -> Option C

4. Quick Check:

   Subclass syntax = colon (:) [OK]

Hint: Use colon (:) to inherit in Swift class declaration [OK]

Common Mistakes:

• Using 'inherits' or 'extends' keywords (not Swift)
• Using wrong symbols like <
• Omitting the colon between subclass and base class

3. Consider this Swift code:

class Vehicle {
    func sound() -> String {
        return "Vroom"
    }
}

class Car: Vehicle {
    override func sound() -> String {
        return "Beep"
    }
}

let myCar = Car()
print(myCar.sound())

What will be printed when this code runs?

medium

A. "Beep"

B. "Vroom"

C. Compilation error

D. "Car"

Solution

1. Step 1: Understand method overriding in subclass

   The subclass Car overrides the sound() method to return "Beep" instead of the base class "Vroom".

2. Step 2: Identify which method is called

   When calling myCar.sound(), the overridden method in Car runs, returning "Beep".

3. Final Answer:

   "Beep" -> Option A

4. Quick Check:

   Overridden method runs = "Beep" [OK]

Hint: Overridden methods in subclass replace base methods [OK]

Common Mistakes:

• Expecting base class method output
• Confusing method override with overload
• Thinking print shows class name

4. Identify the error in this Swift subclass declaration:

class Animal {
    func speak() {
        print("Animal sound")
    }
}

class Cat: Animal {
    func speak() {
        print("Meow")
    }
}

medium

A. Class names must be lowercase

B. Base class cannot have methods

C. Subclass cannot have methods with same name

D. Missing override keyword before speak() in subclass

Solution

1. Step 1: Check method overriding rules in Swift

   When a subclass provides its own version of a base class method, it must use the override keyword.

2. Step 2: Identify missing override keyword

   The speak() method in Cat lacks override, causing a compile error.

3. Final Answer:

   Missing override keyword before speak() in subclass -> Option D

4. Quick Check:

   Override keyword required for method override [OK]

Hint: Use override keyword when redefining base methods [OK]

Common Mistakes:

• Forgetting override keyword
• Thinking subclass methods cannot have same name
• Assuming class names must be lowercase

5. You want to create a base class Employee with a property name and a method work() that prints "Working". Then create a subclass Manager that overrides work() to print "Managing team" and adds a new method holdMeeting() that prints "Meeting started".

Which code correctly implements this?

hard

A. class Employee {
var name: String
init(name: String) {
self.name = name
}
func work() {
print("Working")
}
}

class Manager: Employee {
override func work() {
print("Managing team")
}
func holdMeeting() {
print("Meeting started")
}
}

B. class Employee {
var name: String
func work() {
print("Working")
}
}

class Manager: Employee {
func work() {
print("Managing team")
}
func holdMeeting() {
print("Meeting started")
}
}

C. class Employee {
var name: String
init(name: String) {
self.name = name
}
func work() {
print("Working")
}
}

class Manager: Employee {
func work() {
print("Managing team")
}
func holdMeeting() {
print("Meeting started")
}
}

D. class Employee {
var name: String
init(name: String) {
self.name = name
}
func work() {
print("Working")
}
}

class Manager: Employee {
override func work() {
print("Managing team")
}
}

Solution

1. Step 1: Check base class implementation

   The base class Employee must have a name property with initializer and a work() method printing "Working". class Employee { var name: String init(name: String) { self.name = name } func work() { print("Working") } } class Manager: Employee { override func work() { print("Managing team") } func holdMeeting() { print("Meeting started") } } and C have correct base class with initializer.

2. Step 2: Check subclass overrides and new method

   The subclass Manager must override work() with override keyword and add holdMeeting(). class Employee { var name: String init(name: String) { self.name = name } func work() { print("Working") } } class Manager: Employee { override func work() { print("Managing team") } func holdMeeting() { print("Meeting started") } } correctly uses override and adds holdMeeting(). class Employee { var name: String init(name: String) { self.name = name } func work() { print("Working") } } class Manager: Employee { func work() { print("Managing team") } func holdMeeting() { print("Meeting started") } } misses override. class Employee { var name: String func work() { print("Working") } } class Manager: Employee { func work() { print("Managing team") } func holdMeeting() { print("Meeting started") } } misses initializer in base class. class Employee { var name: String init(name: String) { self.name = name } func work() { print("Working") } } class Manager: Employee { override func work() { print("Managing team") } } misses holdMeeting().

3. Final Answer:

   Option A code correctly implements all requirements -> Option A

4. Quick Check:

   Override + new method + initializer = class Employee { var name: String init(name: String) { self.name = name } func work() { print("Working") } } class Manager: Employee { override func work() { print("Managing team") } func holdMeeting() { print("Meeting started") } } [OK]

Hint: Use override keyword and add new methods in subclass [OK]

Common Mistakes:

• Forgetting override keyword in subclass
• Missing initializer in base class
• Not adding new methods in subclass