Try? for optional result in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to convert the string to an integer.

Swift

let number = Int("123")
print([1])

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Anumber

BInt

Ctry

D123

Attempts:

3 left

💡 Hint

Common Mistakes

Using Int instead of the variable name.

Trying to print try keyword.

Printing the string literal instead of the variable.

The variable number holds the optional integer result from Int("123"). Printing number shows the optional value.

2fill in blank

medium

Complete the code to safely unwrap the optional integer using if let.

Swift

if let value = [1] {
    print("Value is \(value)")
} else {
    print("Conversion failed")
}

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AInt("456")

BInt("abc")

Ctry Int("456")

Dnumber

Attempts:

3 left

💡 Hint

Common Mistakes

Using try Int("456") causes a compile error.

Using Int("abc") which returns nil.

Using an undefined variable number.

Using Int("456") returns an optional integer. The if let safely unwraps it to value.

3fill in blank

hard

Fix the error in the code by completing the try? expression.

Swift

func parseNumber(_ text: String) throws -> Int {
    if let number = Int(text) {
        return number
    } else {
        throw NSError(domain: "ParseError", code: 1)
    }
}

let result = [1] parseNumber("789")
print(result.map(String.init) ?? "No number")

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Atry!

Btry??

Ctry?

Dtry

Attempts:

3 left

💡 Hint

Common Mistakes

Using try without handling errors causes a compile error.

Using try! which crashes if an error is thrown.

Using invalid syntax like try??.

The try? keyword converts the throwing function result into an optional, returning nil if an error occurs.

4fill in blank

hard

Fill both blanks to create a dictionary of word lengths only for words longer than 3 characters.

Swift

let words = ["apple", "cat", "banana", "dog"]
let lengths = [word: [1] for word in words if word.[2] 3]

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Aword.count

Bcount

Ccount >

Dcount <

Attempts:

3 left

💡 Hint

Common Mistakes

Using just count without word. prefix.

Using count < 3 which filters shorter words.

Using word.count but wrong comparison operator.

word.count gets the length of each word. The condition count > 3 filters words longer than 3 characters.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase keys and values only for positive numbers.

Swift

let data = ["a": 1, "b": -2, "c": 3]
let filtered = [[1]: [2] for (key, value) in data where value [3] 0]

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Akey.uppercased()

Bvalue

C>

Dkey

Attempts:

3 left

💡 Hint

Common Mistakes

Using key instead of key.uppercased().

Using value < 0 which filters negative numbers.

Swapping keys and values in the dictionary comprehension.

key.uppercased() converts keys to uppercase. The dictionary includes only entries where value > 0.