Optional chaining with ?. in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to safely access the street name using optional chaining.

Swift

let streetName = person.address[1]street

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

A?.

B!

C.

D->

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.' instead of '?.' causes a compile error if the optional is nil.

Using '!' forces unwrapping and can cause runtime crashes.

Using '?.' allows safe access to street property if address is not nil.

2fill in blank

medium

Complete the code to call the optional method safely using optional chaining.

Swift

let result = person.delegate[1]performAction()

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

A!

B?.

C->

D.

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.' causes compile error if delegate is nil.

Using '!' can cause runtime crash if delegate is nil.

Optional chaining with '?.' safely calls the method if delegate is not nil.

3fill in blank

hard

Fix the error in safely accessing the nested optional property.

Swift

let zipCode = person.address[1]postalCode[2]code

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

A?.

B.

C!

D->

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.' for the first optional causes compile error.

Using '!' can cause runtime crash if optional is nil.

Use optional chaining for the first optional, then normal dot for non-optional property.

4fill in blank

hard

Fill both blanks to safely access the city name from an optional nested property.

Swift

let city = person[1]address[2]city

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

A?.

B.

C!

D->

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.' for the optional property causes errors.

Using '!' risks runtime crashes if the optional is nil.

Use optional chaining for the first optional property, then dot for the non-optional city.

5fill in blank

hard

Fill all three blanks to safely call a method on an optional delegate's optional property.

Swift

let value = person[1]delegate[2]optionalProperty[3]computeValue()

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

A?.

B.

D!

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!' can cause runtime crashes if any optional is nil.

Using '.' for optional properties causes compile errors.

Use optional chaining for optional delegate and optionalProperty, dot for non-optional property.

Practice

1. What does the ?. operator do in Swift when used with optionals?

easy

A. It forces unwraps the optional and crashes if nil.

B. It converts the optional to a non-optional value automatically.

C. It safely accesses a property or method and returns nil if the optional is nil.

D. It assigns a default value if the optional is nil.

Solution

1. Step 1: Understand optional chaining purpose

   The ?. operator is used to safely access properties or methods on an optional value.

2. Step 2: Behavior when optional is nil

   If the optional is nil, the chaining stops and returns nil instead of crashing.

3. Final Answer:

   It safely accesses a property or method and returns nil if the optional is nil. -> Option C

4. Quick Check:

   Optional chaining = safe access with nil return [OK]

Hint: Remember: ? stops if nil, no crash happens [OK]

Common Mistakes:

• Confusing optional chaining with forced unwrapping
• Thinking it unwraps optionals automatically
• Assuming it provides default values

2. Which of the following is the correct syntax to safely access the name property of an optional person object using optional chaining?

easy

A. person->name

B. person!.name

C. person.name?

D. person?.name

Solution

1. Step 1: Identify correct optional chaining syntax

   Optional chaining uses ?. between the optional variable and the property.

2. Step 2: Check each option

   person?.name uses person?.name, which is correct syntax. person!.name forces unwraps with !. person.name? places ? incorrectly. person->name uses invalid operator.

3. Final Answer:

   person?.name -> Option D

4. Quick Check:

   Optional chaining syntax = variable?.property [OK]

Hint: Use ?. right before the property to chain safely [OK]

Common Mistakes:

• Using ! instead of ? for optional chaining
• Placing ? after the property name
• Using invalid operators like ->

3. Consider the following Swift code:

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

let person = Person()
let petName = person.pet?.name
print(petName)

What will be printed?

medium

A. Optional(nil)

B. nil

C. An empty string

D. Runtime error

Solution

1. Step 1: Analyze the optional chaining result

   Since person.pet is nil (not assigned), person.pet?.name returns nil of type String?.

2. Step 2: Understand print output of optional nil

   Printing an optional value that is nil in Swift shows Optional(nil) because the variable itself is an optional.

3. Final Answer:

   Optional(nil) -> Option A

4. Quick Check:

   print(String? nil) = Optional(nil) [OK]

Hint: Printing optional nil shows 'Optional(nil)', not 'nil' [OK]

Common Mistakes:

• Thinking print shows nil
• Expecting empty string output
• Thinking it causes runtime error

4. The following Swift code causes a compile error. What is the mistake?

struct Car {
  var owner: Person?
}

let car = Car()
let ownerName = car.owner?.name!

medium

A. Using forced unwrap (!) after optional chaining without parentheses

B. Missing initialization of car variable

C. Incorrect use of optional chaining operator

D. Person type does not have a name property

Solution

1. Step 1: Identify forced unwrap misuse

   The code uses car.owner?.name! which tries to force unwrap the property without parentheses.

2. Step 2: Correct syntax for forced unwrap after optional chaining

   Forced unwrap must be applied to the entire optional chaining expression with parentheses like (car.owner?.name)!.

3. Final Answer:

   Using forced unwrap (!) after optional chaining without parentheses -> Option A

4. Quick Check:

   Force unwrap after optional chaining needs parentheses [OK]

Hint: Wrap optional chain in () before ! to force unwrap [OK]

Common Mistakes:

• Forgetting parentheses around optional chain before !
• Assuming optional chaining can be forced unwrapped directly
• Ignoring compiler error details

5. Given the following classes:

class Address {
  var city: String?
}
class User {
  var address: Address?
}

let user = User()

Which code safely prints the city name if it exists, or "Unknown" if city or address is nil?

hard

A. print(user.address.city ?? "Unknown")

B. print(user.address?.city ?? "Unknown")

C. print(user.address!.city!)

D. print(user.address?.city!)

Solution

1. Step 1: Understand optional chaining with nil coalescing

   Using user.address?.city safely accesses city, returning nil if address or city is nil.

2. Step 2: Use nil coalescing operator to provide default

   The ?? "Unknown" provides "Unknown" if the optional is nil, avoiding crashes.

3. Final Answer:

   print(user.address?.city ?? "Unknown") -> Option B

4. Quick Check:

   Optional chaining + ?? = safe access with default [OK]

Hint: Combine ?. with ?? to provide default value safely [OK]

Common Mistakes:

• Forcing unwraps causing runtime crashes
• Not handling nil for nested optionals
• Using optional chaining without default fallback