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] }

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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]) }

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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]) })

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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]
}

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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.