Async sequences (AsyncSequence) in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an asynchronous sequence using AsyncSequence protocol.

Swift

struct NumberIterator: AsyncIteratorProtocol {
    typealias Element = Int
    var current = 1
    mutating func next() async -> [1] {
        guard current <= 3 else { return nil }
        defer { current += 1 }
        return current
    }
}

struct NumberSequence: AsyncSequence {
    typealias Element = Int
    func makeAsyncIterator() -> NumberIterator {
        return NumberIterator()
    }
}

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

AInt?

BInt

CString?

DVoid

Attempts:

3 left

💡 Hint

Common Mistakes

Using non-optional return type for next()

Returning a non-async function

The next() method of AsyncIteratorProtocol must return an optional element (Element?) asynchronously. Here, since Element is Int, the return type is Int?.

2fill in blank

medium

Complete the code to asynchronously iterate over the NumberSequence and print each number.

Swift

let sequence = NumberSequence()
Task {
    var iterator = sequence.makeAsyncIterator()
    while let number = await [1] {
        print(number)
    }
}

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

Aiterator.next()

Bsequence.next()

Citerator.next

Dsequence.makeAsyncIterator()

Attempts:

3 left

💡 Hint

Common Mistakes

Calling next() on the sequence instead of the iterator

Forgetting to use await

To get the next element asynchronously, call await iterator.next(). This returns the next optional element from the iterator.

3fill in blank

hard

Fix the error in the async for loop to iterate over the sequence correctly.

Swift

let sequence = NumberSequence()
Task {
    for await [1] in sequence {
        print([1])
    }
}

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

Avalue

Bnum

Citem

Dnumber

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent variable names inside the loop

Using reserved keywords as variable names

The variable name inside the loop can be any valid identifier. Here, number is a clear and descriptive name for each element in the sequence.

4fill in blank

hard

Fill both blanks to create a filtered async sequence that yields only even numbers.

Swift

struct EvenNumberSequence: AsyncSequence {
    typealias Element = Int
    let baseSequence: NumberSequence
    func makeAsyncIterator() -> Iterator {
        return Iterator(baseIterator: baseSequence.makeAsyncIterator() as! NumberIterator)
    }
    struct Iterator: AsyncIteratorProtocol {
        typealias Element = Int
        var baseIterator: NumberIterator
        mutating func next() async -> [1] {
            while let number = await baseIterator.next() {
                if number [2] 2 == 0 {
                    return number
                }
            }
            return nil
        }
    }
}

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

AInt?

BInt

C%

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using non-optional return type for next()

Using equality operator instead of modulo for even check

The next() method returns an optional Int (Int?). To check if a number is even, use the modulo operator % to test if the remainder when divided by 2 is zero.

5fill in blank

hard

Fill all three blanks to create a dictionary from an async sequence where keys are string versions of numbers and values are the numbers themselves, filtering only numbers greater than 1.

Swift

func createDictionary(from sequence: NumberSequence) async -> [String: Int] {
    var result = [String: Int]()
    var iterator = sequence.makeAsyncIterator()
    while let number = await [1] {
        if number [2] 1 {
            result[[3]] = number
        }
    }
    return result
}

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

Aiterator.next()

B>

CString(number)

Dnumber.next()

Attempts:

3 left

💡 Hint

Common Mistakes

Calling next() on the wrong variable

Using wrong comparison operator

Not converting number to string for dictionary key

To get the next element asynchronously, call await iterator.next(). To filter numbers greater than 1, use >. To use the number as a key in the dictionary, convert it to a string with String(number).