Optional declaration with ? suffix in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Swift code with optional?

Consider the following Swift code. What will be printed when it runs?

Swift

var name: String? = "Alice"
print(name)

A Optional("Alice")

B Alice

C nil

D Error: Variable 'name' not initialized

💡 Hint

Remember that printing an optional variable shows it wrapped in Optional(...) if it has a value.

In Swift, an optional variable holds either a value or nil. Printing an optional directly shows Optional(value) if it has a value.

❓ Predict Output

intermediate

What is the value of 'age' after this code runs?

Look at this Swift code snippet. What is the value of 'age' after execution?

Swift

var age: Int? = nil
age = 30

A Optional(30)

B nil

C 30

D Error: Cannot assign non-optional to optional

💡 Hint

Assigning a non-optional value to an optional variable wraps it automatically.

Assigning 30 to 'age' wraps it as Optional(30). The variable 'age' is still optional but holds the value 30.

❓ Predict Output

advanced

What error does this code produce?

What error will this Swift code cause when compiled?

Swift

var score: Int? = 10
var total: Int = score + 5

A Syntax error: Missing optional unwrapping

B No error, total is 15

C Runtime error: nil value unwrapped

D Type error: Cannot add 'Int?' and 'Int'

💡 Hint

You cannot directly add an optional Int and a non-optional Int without unwrapping.

The compiler does not allow adding an optional Int and an Int directly. You must unwrap the optional first.

🧠 Conceptual

advanced

Which statement about optional declaration with '?' is true?

Choose the correct statement about declaring variables with '?' in Swift.

A Variables with '?' cannot be assigned nil after initialization.

B A variable declared with '?' must always have a value assigned.

C A variable declared with '?' can hold a value or nil.

D Using '?' means the variable is a constant and cannot change.

💡 Hint

Think about what '?' means in Swift variable declarations.

The '?' suffix means the variable is optional and can hold a value or nil.

❓ Predict Output

expert

What is the output of this Swift optional chaining code?

Analyze the code and select the output printed.

Swift

class Person {
    var pet: Pet?
}

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

let person = Person()
print(person.pet?.name ?? "No pet")

A Optional(nil)

B No pet

C nil

D Runtime error: Unexpected nil

💡 Hint

Optional chaining returns nil if any part is nil, and the nil-coalescing operator provides a default.

Since 'person.pet' is nil, 'person.pet?.name' is nil, so the default "No pet" is printed.

Practice

1. What does adding a ? after a type in Swift mean?

easy

A. The variable can hold a value or be nil (missing).

B. The variable is a constant and cannot change.

C. The variable must always have a value.

D. The variable is an array of that type.

Solution

1. Step 1: Understand the meaning of ? in Swift type declaration

   In Swift, adding ? after a type means the variable is optional and can hold either a value or nil.

2. Step 2: Compare with other options

   Other options describe constants, non-optional variables, or arrays, which are unrelated to the ? suffix.

3. Final Answer:

   The variable can hold a value or be nil (missing). -> Option A

4. Quick Check:

   Optional type = value or nil [OK]

Hint: Remember: ? means value can be missing [OK]

Common Mistakes:

• Thinking ? makes a constant
• Confusing optional with array type
• Assuming optional must always have a value

2. Which of these is the correct way to declare an optional integer in Swift?

easy

A. var number: Int?

B. var number: ?Int

C. var number: Int!

D. var number: OptionalInt

Solution

1. Step 1: Recall Swift syntax for optional declaration

   In Swift, an optional type is declared by placing ? after the type name, like Int?.

2. Step 2: Check each option

   var number: Int? uses Int? correctly. var number: ?Int places ? before the type, which is invalid. var number: Int! uses ! which is an implicitly unwrapped optional, different from ?. var number: OptionalInt uses a non-existent type OptionalInt.

3. Final Answer:

   var number: Int? -> Option A

4. Quick Check:

   Optional syntax = type followed by ? [OK]

Hint: Put ? right after the type name [OK]

Common Mistakes:

• Placing ? before the type
• Confusing ? with !
• Using non-existent types like OptionalInt

3. What will be the output of this Swift code?

var name: String? = "Alice"
print(name)

medium

A. nil

B. Alice

C. Optional("Alice")

D. Error: Cannot print optional directly

Solution

1. Step 1: Understand printing an optional variable

   When printing an optional variable directly, Swift shows it wrapped as Optional(value) if it has a value.

2. Step 2: Check the variable value

   The variable name holds the string "Alice", so printing it shows Optional("Alice").

3. Final Answer:

   Optional("Alice") -> Option C

4. Quick Check:

   Printing optional shows Optional(value) [OK]

Hint: Printing optional shows Optional(...) wrapper [OK]

Common Mistakes:

• Expecting direct value without Optional()
• Assuming printing optional causes error
• Confusing nil with a string value

4. Identify the error in this Swift code:

var age: Int?
age = 25
print(age + 5)

medium

A. Variable age is not initialized.

B. Optional variables cannot be printed.

C. Missing semicolon after age = 25.

D. Cannot add Int? and Int directly.

Solution

1. Step 1: Analyze the operation on optional and Int

   The variable age is an optional Int (Int?), but adding it directly to an Int (5) is not allowed without unwrapping.

2. Step 2: Understand Swift's type safety

   Swift requires unwrapping optionals before using them in arithmetic operations to avoid runtime errors.

3. Final Answer:

   Cannot add Int? and Int directly. -> Option D

4. Quick Check:

   Optional must be unwrapped before arithmetic [OK]

Hint: Unwrap optionals before arithmetic operations [OK]

Common Mistakes:

• Trying to add optional and non-optional directly
• Ignoring need to unwrap optionals
• Thinking semicolons are required in Swift

5. Given this Swift code, how can you safely print the length of the optional string text?

var text: String? = "Hello"
// Your code here

hard

A. print(text ?? 0)

B. if let safeText = text { print(safeText.count) } else { print(0) }

C. print(text!)

D. print(text.count)

Solution

1. Step 1: Understand optional unwrapping with if let

   Using if let safely unwraps the optional text only if it has a value, avoiding crashes.

2. Step 2: Use unwrapped value to get length

   Inside the if let block, safeText.count gives the length. If text is nil, print 0.

3. Step 3: Check other options

   print(text.count) tries to access count on optional directly (error). print(text!) force unwraps without check (unsafe). print(text ?? 0) uses nil coalescing but mismatches types.

4. Final Answer:

   if let safeText = text { print(safeText.count) } else { print(0) } -> Option B

5. Quick Check:

   Use if let to safely unwrap optionals [OK]

Hint: Use if let to unwrap optionals safely [OK]

Common Mistakes:

• Forcing unwrap without checking nil
• Accessing properties on optional without unwrapping
• Using nil coalescing with wrong types