Try! for forced unwrap in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to force unwrap the result of the throwing function using try!

Swift

func getNumber() throws -> Int {
    return 42
}

let number = [1] getNumber()

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

Atry?

Btry

Ctry!

Dcatch

Attempts:

3 left

💡 Hint

Common Mistakes

Using try without handling errors causes a compile error.

Using try? returns an optional, not a forced unwrap.

Using catch is for error handling, not unwrapping.

Using try! forces the program to unwrap the result of a throwing function. If the function throws an error, the program will crash.

2fill in blank

medium

Complete the code to force unwrap the string returned by the throwing function using try!

Swift

func fetchName() throws -> String {
    return "Alice"
}

let name: String = [1] fetchName()

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Acatch

Btry?

Ctry

Dtry!

Attempts:

3 left

💡 Hint

Common Mistakes

Using try? returns an optional String, not a forced unwrap.

Using try without a do-catch block causes an error.

Using catch is not valid syntax here.

try! unwraps the result directly and crashes if an error is thrown.

3fill in blank

hard

Fix the error by force unwrapping the result of the throwing function using try!

Swift

func getValue() throws -> Double {
    return 3.14
}

let value = [1] getValue()

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

Atry

Btry!

Ctry?

Dcatch

Attempts:

3 left

💡 Hint

Common Mistakes

Using try without a do-catch block causes a compile error.

Using try? returns an optional, not a forced unwrap.

Using catch is not valid here.

Using try! forces the unwrap of the throwing function's result, fixing the error.

4fill in blank

hard

Fill both blanks to populate the dictionary that stores the forced unwrapped length of each word if the length is greater than 3.

Swift

func getLength(_ word: String) throws -> Int {
    return word.count
}

let words = ["apple", "cat", "banana", "dog"]
var lengths = [String: Int]()
for word in words {
    if [1](word).count [2] 3 {
        lengths[word] = try! getLength(word)
    }
}

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AString

B>

C<

DInt

Attempts:

3 left

💡 Hint

Common Mistakes

Using Int(word) instead of String(word) causes errors.

Using < instead of > changes the logic.

Not using try! to force unwrap the function result.

We use String(word).count to get the length of the word and check if it is greater than 3.

5fill in blank

hard

Fill both blanks to create a dictionary that maps uppercase words to their forced unwrapped lengths if the length is greater than 4.

Swift

func getLength(_ word: String) throws -> Int {
    return word.count
}

let words = ["apple", "pear", "banana", "kiwi"]
var lengths = [String: Int]()
for word in words {
    if word.count [1] 4 {
        lengths[word[2]] = try! getLength(word)
    }
}

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A>

B.uppercased()

C<

D.lowercased()

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of > changes the condition.

Using .lowercased() instead of .uppercased() changes the key.

Not using try! to force unwrap the function result.

The code checks if the word length is greater than 4, then uses .uppercased() to make the key uppercase and stores the forced unwrapped length.