Force unwrapping with ! and its danger in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to force unwrap the optional value.

Swift

let name: String? = "Alice"
let unwrappedName = name[1]

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

A?

B!

C.

D:

Attempts:

3 left

💡 Hint

Common Mistakes

Using '?' instead of '!' which only safely unwraps.

Using '.' which accesses properties or methods.

Using ':' which is not related to unwrapping.

The exclamation mark (!) is used to force unwrap an optional in Swift, extracting the value inside.

2fill in blank

medium

Complete the code to safely unwrap the optional using if-let instead of force unwrapping.

Swift

let age: Int? = 30
if let unwrappedAge = [1] {
    print("Age is \(unwrappedAge)")
}

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

Aage

Bage!

Cage?

Dage ?? 0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'age!' inside if-let which causes a crash if nil.

Using 'age?' which is not valid syntax here.

Using 'age ?? 0' which provides a default but is not unwrapping.

In if-let, you unwrap the optional by just using its name without force unwrapping.

3fill in blank

hard

Fix the error by replacing force unwrapping with safe unwrapping.

Swift

let city: String? = nil
print("City is \(city[1])")

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

A!

B? ?? "Unknown"

C?

D?? "Unknown"

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!' causes a runtime crash if city is nil.

Using '?' alone does not provide a default value.

Using '? ??' is invalid syntax.

Using '??' provides a default value if the optional is nil, avoiding a crash.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that includes only names with length greater than 3.

Swift

let names = ["Tom", "Anna", "Bob", "Catherine"]
let filteredNames = names.filter { $0 [1] 3 }.map { $0[2] }

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

A>

B!

C.uppercased()

D.lowercased()

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!' in filter condition causes syntax error.

Using '.lowercased()' instead of '.uppercased()' changes output.

Using '<' filters wrong names.

We filter names with length greater than 3 and then convert them to uppercase.

5fill in blank

hard

Fill all three blanks to create a dictionary of names and their lengths, including only names longer than 4 characters.

Swift

let names = ["John", "Elizabeth", "Mark", "Samantha"]
let nameLengths = Dictionary(uniqueKeysWithValues: names.filter { $0.count [1] 4 }.map { ([2], [3]) in ([2], [3].count) })

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

A>

Bname

Cword

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' filters wrong names.

Using different variable names inconsistently.

Using force unwrapping inside map.

We filter names longer than 4, then map each name to a tuple of (name, name.count).