Await for calling async functions in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this async function call?

Consider the following Swift code using async/await. What will be printed when running this code?

Swift

func fetchNumber() async -> Int {
    return 42
}

func printNumber() async {
    let number = await fetchNumber()
    print("Number is \(number)")
}

Task {
    await printNumber()
}

A Number is 42

B Number is Optional(42)

C Compilation error due to missing await

D Runtime error: Cannot call async function without await

💡 Hint

Remember to use await when calling async functions to get their result.

The function fetchNumber returns 42 asynchronously. The printNumber function awaits fetchNumber and prints the result. The output is 'Number is 42'.

❓ Predict Output

intermediate

What happens if you omit await when calling an async function?

Look at this Swift code snippet. What error or output will it produce?

Swift

func fetchText() async -> String {
    return "Hello"
}

func test() {
    let text = fetchText()
    print(text)
}

test()

A Runtime error: Async function called without await

B Prints: Hello

C Prints: Optional("Hello")

D Compilation error: Call to async function in a synchronous context without await

💡 Hint

Async functions must be awaited or called from async context.

Calling an async function without await in a synchronous function causes a compilation error in Swift.

🔧 Debug

advanced

Why does this async function not print the expected result?

Examine the code below. It is supposed to print "Result: 10" but prints nothing. What is the reason?

Swift

func calculate() async -> Int {
    return 10
}

func run() {
    Task {
        let value = await calculate()
        print("Result: \(value)")
    }
}

run()

A print statement syntax is incorrect

B run() is synchronous and exits before Task completes, so print never happens

C Missing await before Task, so Task is not executed

D calculate() is not awaited properly inside Task

💡 Hint

Consider the lifecycle of asynchronous tasks and the main thread.

The run() function starts a Task but returns immediately. The program may exit before the Task finishes, so print is never seen.

📝 Syntax

advanced

Which option correctly awaits an async function inside a synchronous function?

Given an async function fetchData(), which code snippet correctly calls it from a synchronous function?

Swift

func fetchData() async -> String {
    return "Data"
}

func syncCaller() {
    // Which option is correct here?
}

A

Task {
    let data = await fetchData()
    print(data)
}

B let data = await fetchData()

C fetchData().wait()

D let data = fetchData()

💡 Hint

You cannot use await directly in synchronous functions, but you can start a Task.

Option A uses Task to run async code inside a synchronous function. Option A is invalid syntax, C is not a Swift method, D ignores async nature.

🚀 Application

expert

How many times will the print statement execute in this async sequence?

Analyze the code below. How many times will "Done" be printed?

Swift

func delayedPrint() async {
    for i in 1...3 {
        await Task.sleep(UInt64(500_000_000)) // 0.5 seconds
        print("Done")
    }
}

Task {
    await delayedPrint()
}

A 0 times

B 1 time

C 3 times

D Infinite times

💡 Hint

The loop runs 3 times with a delay before each print.

The for loop runs 3 times, each time awaiting 0.5 seconds then printing "Done". So output is printed 3 times.