Classes are reference types in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of class instance assignment

What is the output of this Swift code?

Swift

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

let person1 = Person(name: "Alice")
let person2 = person1
person2.name = "Bob"
print(person1.name)

A "Alice"

B "Bob"

C Compilation error

D "person1"

💡 Hint

Remember that classes are reference types, so both variables point to the same instance.

In Swift, classes are reference types. Assigning person1 to person2 means both refer to the same object. Changing person2.name changes the name for person1 as well.

❓ Predict Output

intermediate

Effect of modifying a class property via one reference

What will be printed by this Swift code?

Swift

class Counter {
    var count = 0
}

let c1 = Counter()
let c2 = c1
c2.count += 5
print(c1.count)

A 5

B 0

C Compilation error

D nil

💡 Hint

Both variables refer to the same instance, so changes via one affect the other.

Since Counter is a class, c1 and c2 point to the same object. Increasing c2.count changes c1.count as well.

🔧 Debug

advanced

Why does this code print 20 instead of 5?

Consider this Swift code. Why does it print 20 instead of 5?

Swift

class Number {
    var value: Int
    init(value: Int) {
        self.value = value
    }
}

func doubleValue(_ num: Number) {
    num.value *= 2
}

let n = Number(value: 5)
doubleValue(n)
doubleValue(n)
print(n.value)

A Because the function doubles the value once, but the print statement is wrong

B Because the function doubles the value twice, so 5 * 2 = 10

C Because classes are reference types, the same instance is modified twice, resulting in 20

D Because the function doubles the same instance twice, so 5 * 2 * 2 = 20

💡 Hint

Trace the value changes step by step.

The function doubleValue doubles the value property each time it is called. It is called twice, so starting from 5: first call makes it 10, second call makes it 20. The print statement prints 20.

Option C correctly explains that classes are reference types, so the same instance is modified twice.

❓ Predict Output

advanced

Output when copying class instance to a new variable

What is the output of this Swift code?

Swift

class Box {
    var content: String
    init(content: String) {
        self.content = content
    }
}

var box1 = Box(content: "Apple")
var box2 = box1
box2.content = "Orange"
print(box1.content)

A "Orange"

B Runtime error

C "box1"

D "Apple"

💡 Hint

Remember that assigning a class instance copies the reference, not the object.

Both box1 and box2 refer to the same Box instance. Changing box2.content changes box1.content as well.

❓ Predict Output

expert

Output of modifying nested class references

What will this Swift code print?

Swift

class Engine {
    var horsepower: Int
    init(horsepower: Int) {
        self.horsepower = horsepower
    }
}

class Car {
    var engine: Engine
    init(engine: Engine) {
        self.engine = engine
    }
}

let engine1 = Engine(horsepower: 150)
let car1 = Car(engine: engine1)
let car2 = car1
car2.engine.horsepower = 200
print(car1.engine.horsepower)

A 150

B nil

C Compilation error

D 200

💡 Hint

Both cars share the same engine instance because classes are reference types.

Both car1 and car2 refer to the same Car instance. The engine property is also a class instance shared by both. Changing car2.engine.horsepower changes car1.engine.horsepower.

Practice

1. What happens when you assign an instance of a class to another variable in Swift?

easy

A. A new copy of the object is created.

B. Both variables refer to the same object in memory.

C. The original object is deleted.

D. The variables become constants automatically.

Solution

1. Step 1: Understand class assignment behavior

   In Swift, classes are reference types, so assigning one instance to another variable copies the reference, not the object.

2. Step 2: Effect of assignment on variables

   Both variables point to the same memory location, so changes via one variable affect the other.

3. Final Answer:

   Both variables refer to the same object in memory. -> Option B

4. Quick Check:

   Classes are reference types = A [OK]

Hint: Class assignment copies reference, not object [OK]

Common Mistakes:

• Thinking a new object is created on assignment
• Confusing classes with structs (value types)
• Assuming variables become constants automatically

2. Which of the following is the correct way to declare a class in Swift?

easy

A. class Dog: String { var name }

B. struct Dog { var name: String }

C. class Dog { var name: String }

D. func Dog() { var name: String }

Solution

1. Step 1: Identify correct class syntax

   In Swift, a class is declared using the keyword 'class' followed by the class name and curly braces containing properties.

2. Step 2: Check each option

   class Dog { var name: String } correctly declares a class with a property. struct Dog { var name: String } declares a struct, not a class. class Dog: String { var name } has invalid syntax with inheritance from String and missing property type. func Dog() { var name: String } uses 'func' which is for functions, not classes.

3. Final Answer:

   class Dog { var name: String } -> Option C

4. Quick Check:

   Correct class syntax = A [OK]

Hint: Use 'class' keyword and braces for class declaration [OK]

Common Mistakes:

• Using 'struct' instead of 'class'
• Incorrect inheritance syntax
• Using 'func' instead of 'class'

3. What will be printed by the following Swift code?

class Counter {
  var count = 0
}

let a = Counter()
let b = a
b.count = 5
print(a.count)

medium

A. 0

B. Compilation error

C. nil

D. 5

Solution

1. Step 1: Understand reference assignment

   Variable 'b' is assigned to 'a', so both refer to the same Counter instance.

2. Step 2: Effect of changing 'b.count'

   Changing 'b.count' to 5 changes the shared object's property, so 'a.count' also becomes 5.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Class references share data = 5 [OK]

Hint: Changing one reference changes all references [OK]

Common Mistakes:

• Assuming 'a.count' stays 0 after changing 'b.count'
• Expecting a compilation error
• Confusing with value type behavior

4. Identify the error in this Swift code snippet:

class Person {
  var name: String
}

let p1 = Person()
p1.name = "Alice"

medium

A. Class Person cannot be instantiated with 'let'.

B. Cannot assign value to 'name' because it is a constant.

C. Missing initializer for 'name' property in class Person.

D. No error; code runs fine.

Solution

1. Step 1: Check class property initialization

   The class Person has a non-optional property 'name' without an initial value or initializer.

2. Step 2: Understand Swift's requirement for property initialization

   Swift requires all non-optional properties to be initialized before use, either with a default value or in an initializer.

3. Final Answer:

   Missing initializer for 'name' property in class Person. -> Option A

4. Quick Check:

   Non-optional properties need initialization = C [OK]

Hint: Non-optional properties must be initialized [OK]

Common Mistakes:

• Thinking 'let' prevents property assignment
• Assuming classes can't be instantiated with 'let'
• Ignoring property initialization rules

5. You want to create a shared settings object that multiple parts of your app can modify and see updates immediately. Which approach best uses Swift classes to achieve this?

hard

A. Create a single instance of a Settings class and pass its reference around.

B. Create multiple instances of a Settings struct and copy them as needed.

C. Use global variables instead of classes for shared data.

D. Use a function that returns a new Settings object every time.

Solution

1. Step 1: Understand shared mutable state with classes

   Classes are reference types, so a single instance shared by multiple parts allows changes to be seen everywhere.

2. Step 2: Evaluate options for shared settings

   Create a single instance of a Settings class and pass its reference around. uses one shared instance, which fits the requirement. Create multiple instances of a Settings struct and copy them as needed. uses structs (value types), so copies won't reflect changes. Use global variables instead of classes for shared data. uses globals, which is not recommended for maintainability. Use a function that returns a new Settings object every time. creates new objects each time, so no shared state.

3. Final Answer:

   Create a single instance of a Settings class and pass its reference around. -> Option A

4. Quick Check:

   Shared mutable object = B [OK]

Hint: Use one shared class instance for shared mutable data [OK]

Common Mistakes:

• Using structs which copy data instead of sharing
• Relying on global variables for sharing
• Creating new instances instead of sharing one