CompactMap for optional unwrapping in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to unwrap optionals using compactMap.

Swift

let numbers: [Int?] = [1, nil, 3, nil, 5]
let unwrapped = numbers.[1] { $0 }

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

AcompactMap

Bmap

Cfilter

DflatMap

Attempts:

3 left

💡 Hint

Common Mistakes

Using map keeps nil values, resulting in an array of optionals.

Using filter only filters but does not unwrap optionals.

The compactMap method unwraps optionals and removes nil values from the array.

2fill in blank

medium

Complete the code to unwrap and double the numbers using compactMap.

Swift

let numbers: [Int?] = [2, nil, 4, nil, 6]
let doubled = numbers.compactMap { [1] }

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

A$0 ?? 0

B$0 + 2

C$0

D$0 * 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using $0 ?? 0 returns 0 for nils, which compactMap does not require.

Using map instead of compactMap keeps nils.

The closure unwraps each optional and doubles the value. compactMap removes nils.

3fill in blank

hard

Fix the error in the code to unwrap optionals correctly with compactMap.

Swift

let strings: [String?] = ["1", nil, "3", "four"]
let numbers = strings.compactMap { Int([1]) }

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

AString($0)

B$0

C$0 ?? "0"

DInt($0)

Attempts:

3 left

💡 Hint

Common Mistakes

Using $0 ?? "0" converts nils to "0", which changes the logic.

Using String($0) is redundant.

Option D creates Int(Int($0)), causing a compiler error.

The closure receives a String?. Pass $0 to Int(_:), which safely returns Int? for compactMap to unwrap non-nil values.

4fill in blank

hard

Fill both blanks to create a dictionary of word lengths, ignoring short words.

Swift

let words = ["apple", "to", "banana", "go"]
let lengths = Dictionary(uniqueKeysWithValues: words.filter { $0.count > 2 }.map { ([1], [2]) })

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

Aword

Bword.count

Clen(word)

Dword.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(word) or word.length which are not valid in Swift.

Using word.count in the key position instead of the value.

The dictionary uses the word as key and its count as value, filtering words longer than 2.

5fill in blank

hard

Fill all three blanks to filter and transform optional strings to uppercase.

Swift

let optionalStrings: [String?] = ["cat", nil, "dog", ""]
let result = optionalStrings.compactMap { [1] in
    guard let str = [2], !str.isEmpty else { return nil }
    return [3]
}

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

Astr

B$0

Cstr.uppercased()

DoptionalStrings

Attempts:

3 left

💡 Hint

Common Mistakes

Using $0 as closure parameter causes syntax error here.

Returning the original string instead of uppercase.

Not unwrapping the optional properly.

The closure parameter is named str (optional string). Guard unwraps str (shadowing the parameter), checks non-empty, returns uppercase.

Practice

1. What does the compactMap function do when used on an array of optionals in Swift?

easy

A. It unwraps optionals and removes all nil values from the array.

B. It converts all values to strings regardless of their type.

C. It sorts the array in ascending order.

D. It replaces all nil values with zero.

Solution

1. Step 1: Understand what compactMap does

   compactMap takes an array of optionals and unwraps each element, removing any nil values.

2. Step 2: Compare with other options

   It does not convert types, sort, or replace nil with zero. It only filters out nil and unwraps values.

3. Final Answer:

   It unwraps optionals and removes all nil values from the array. -> Option A

4. Quick Check:

   compactMap removes nil = B [OK]

Hint: Remember: compactMap unwraps and drops nils [OK]

Common Mistakes:

• Confusing compactMap with map (which keeps nils)
• Thinking compactMap sorts or changes types
• Assuming compactMap replaces nil with default values

2. Which of the following is the correct syntax to use compactMap on an array numbers of type [Int?] to get an array of non-optional Int values?

easy

A. let result = numbers.compactMap { $0 }

B. let result = numbers.compactMap { Int($0) }

C. let result = numbers.compactMap { $0! }

D. let result = numbers.compactMap { String($0) }

Solution

1. Step 1: Understand the input and output types

   numbers is an array of optional integers [Int?]. We want to unwrap and remove nil values to get [Int].

2. Step 2: Check each option's correctness

   let result = numbers.compactMap { $0 } uses compactMap { $0 } which unwraps optionals and removes nil. let result = numbers.compactMap { Int($0) } tries to convert optional Int to Int again, which is unnecessary. let result = numbers.compactMap { $0! } force unwraps, which can crash if nil. let result = numbers.compactMap { String($0) } converts to String, changing type.

3. Final Answer:

   let result = numbers.compactMap { $0 } -> Option A

4. Quick Check:

   Use compactMap with identity closure to unwrap optionals [OK]

Hint: Use compactMap { $0 } to unwrap optionals safely [OK]

Common Mistakes:

• Using force unwrap inside compactMap causing runtime errors
• Trying to convert optionals unnecessarily
• Changing the type unintentionally

3. What is the output of the following Swift code?

let values: [Int?] = [1, nil, 3, nil, 5]
let unwrapped = values.compactMap { $0 }
print(unwrapped)

medium

A. Compilation error

B. [1, 3, 5]

C. [1, nil, 3, nil, 5]

D. [Optional(1), Optional(3), Optional(5)]

Solution

1. Step 1: Understand the input array

   The array values contains optional integers with some nil values.

2. Step 2: Apply compactMap

   compactMap { $0 } unwraps optionals and removes nil values, resulting in [1, 3, 5].

3. Final Answer:

   [1, 3, 5] -> Option B

4. Quick Check:

   compactMap removes nils and unwraps = [1, 3, 5] [OK]

Hint: compactMap removes nils and unwraps optionals [OK]

Common Mistakes:

• Expecting optionals in output instead of unwrapped values
• Thinking nils remain in the output
• Confusing compactMap with map

4. Identify the error in this Swift code snippet and choose the correct fix:

let strings: [String?] = ["apple", nil, "banana"]
let lengths = strings.compactMap { $0.count }
print(lengths)

medium

A. Error: Use force unwrap $0! before accessing count

B. Error: compactMap should be replaced with map

C. No error; code runs correctly

D. Error: Cannot use $0.count on optional; fix by using compactMap { $0?.count }

Solution

1. Step 1: Analyze the closure inside compactMap

   The closure uses $0.count, but $0 is an optional String?. Accessing count directly on optional causes error.

2. Step 2: Correct the closure to safely unwrap

   Using $0?.count returns an optional Int, which compactMap unwraps and removes nils, fixing the error.

3. Final Answer:

   Error: Cannot use $0.count on optional; fix by using compactMap { $0?.count } -> Option D

4. Quick Check:

   Use optional chaining inside compactMap closure [OK]

Hint: Use optional chaining ($0?.property) inside compactMap [OK]

Common Mistakes:

• Trying to access property directly on optional causing error
• Using map instead of compactMap losing nil filtering
• Force unwrapping causing runtime crashes

5. Given a dictionary [String: Int?] representing item stock counts, how can you create a new dictionary with only items that have a non-nil stock count using compactMap or similar methods?

hard

A. Use filter { $0.value != nil } without unwrapping values.

B. Use mapValues { $0! } to force unwrap all values.

C. Use compactMapValues { $0 } to unwrap and filter out nil values.

D. Use compactMap { $0 } directly on the dictionary.

Solution

1. Step 1: Understand dictionary optional values

   The dictionary has optional Int values. We want to remove keys with nil values and unwrap the rest.

2. Step 2: Use compactMapValues to unwrap and filter

   compactMapValues { $0 } unwraps optionals in values and removes entries with nil values, producing a dictionary with non-optional Int values.

3. Final Answer:

   Use compactMapValues { $0 } to unwrap and filter out nil values. -> Option C

4. Quick Check:

   compactMapValues unwraps and filters dictionary values [OK]

Hint: Use compactMapValues to unwrap and remove nil dictionary values [OK]

Common Mistakes:

• Using mapValues with force unwrap risking crashes
• Filtering without unwrapping values
• Trying to use compactMap directly on dictionary