Guard let for early exit in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Swift code using guard let?

Consider the following Swift function that uses guard let for early exit. What will be printed when calling checkName(nil)?

Swift

func checkName(_ name: String?) {
    guard let unwrappedName = name else {
        print("Name is missing")
        return
    }
    print("Hello, \(unwrappedName)!")
}

checkName(nil)

A Hello, nil!

B Compilation error

C Name is missing

D Hello, !

💡 Hint

Think about what happens when the optional is nil and how guard let handles it.

The guard statement checks if 'name' is non-nil. Since 'name' is nil, the else block runs, printing 'Name is missing' and returning early.

❓ Predict Output

intermediate

What does this Swift function print when given a non-nil value?

Analyze this Swift function using guard let. What will be printed when calling checkAge(25)?

Swift

func checkAge(_ age: Int?) {
    guard let validAge = age, validAge >= 18 else {
        print("Underage or no age provided")
        return
    }
    print("Access granted for age \(validAge)")
}

checkAge(25)

A Underage or no age provided

B Access granted for age 25

C Access granted for age nil

D Runtime error

💡 Hint

Check the conditions inside the guard statement carefully.

The guard unwraps 'age' and checks if it is 18 or older. Since 25 meets both conditions, the else block is skipped and the success message prints.

🔧 Debug

advanced

Identify the error in this Swift code using guard let

This Swift function intends to unwrap an optional string and print it. What error will occur when compiling or running this code?

Swift

func printMessage(_ message: String?) {
    guard let message = message else {
        print("No message to print")
        return
    }
    print(message)
}

printMessage(nil)

A Prints 'No message to print' only

B Runtime error: Force unwrap nil optional

C Prints 'No message to print' and then crashes

D Compilation error: Missing return in guard else block

💡 Hint

Remember what guard else block must always do.

The guard else block must exit the current scope (return, break, continue, or throw). Here it only prints but does not return, causing a compilation error.

❓ Predict Output

advanced

What is the output of this nested guard let example?

Consider this Swift function with nested guard let statements. What will be printed when calling processData(name: "Alice", age: nil)?

Swift

func processData(name: String?, age: Int?) {
    guard let validName = name else {
        print("Name missing")
        return
    }
    guard let validAge = age else {
        print("Age missing")
        return
    }
    print("Name: \(validName), Age: \(validAge)")
}

processData(name: "Alice", age: nil)

A Age missing

B Compilation error

C Name: Alice, Age: nil

D Name missing

💡 Hint

Check which guard fails first and what it prints.

The first guard unwraps 'name' successfully. The second guard fails because 'age' is nil, so it prints 'Age missing' and returns early.

🧠 Conceptual

expert

Why is guard let preferred for early exit in Swift?

Which of the following best explains why Swift developers use guard let for early exit instead of nested if-let statements?

A Guard let allows unwrapping optionals and keeps the main code path less nested and more readable by exiting early on failure.

B Guard let automatically unwraps optionals without needing else blocks or returns.

C Guard let is faster at runtime than if-let because it uses compiler optimizations.

D Guard let can only be used with non-optional values, making code safer.

💡 Hint

Think about code readability and flow control.

Guard let requires an else block that exits early, which avoids deep nesting and keeps the main logic clear and linear.