Existential types (any keyword) in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of existential type usage with 'any' keyword

What is the output of this Swift code using the 'any' keyword for existential types?

Swift

protocol Greetable {
    func greet() -> String
}

struct Person: Greetable {
    func greet() -> String {
        "Hello from Person"
    }
}

func welcome(_ greeter: any Greetable) {
    print(greeter.greet())
}

let p = Person()
welcome(p)

A Protocol 'Greetable' cannot be used as a type

B Error: Missing 'any' keyword before 'Greetable'

C Hello from Person

D Hello from any

💡 Hint

Remember that 'any' keyword is used to explicitly mark existential types in Swift 5.7+.

The 'any' keyword allows the protocol to be used as a type. The function calls the greet() method on the Person instance, printing 'Hello from Person'.

❓ Predict Output

intermediate

Value stored in variable with 'any' existential type

What is the value of variable 'message' after running this Swift code?

Swift

protocol Describable {
    func describe() -> String
}

struct Animal: Describable {
    func describe() -> String {
        "I am an animal"
    }
}

let creature: any Describable = Animal()
let message = creature.describe()

A "I am a creature"

B "I am an animal"

C Compile-time error: protocol used without 'any'

D Runtime error: cannot call describe() on existential

💡 Hint

The 'any' keyword allows calling protocol methods on the existential value.

The variable 'creature' is typed as 'any Describable', so calling describe() returns the string from Animal's implementation.

🔧 Debug

advanced

Identify the error with existential type usage

What error does this Swift code produce?

Swift

protocol Runnable {
    func run()
}

func execute(task: Runnable) {
    task.run()
}

struct Job: Runnable {
    func run() {
        print("Job running")
    }
}

let job = Job()
execute(task: job)

A No error, prints 'Job running'

B Error: Missing 'any' keyword before 'Runnable'

C Error: Protocol 'Runnable' can only be used as a generic constraint because it has Self or associated type requirements

D Runtime error: cannot call run() on protocol type

💡 Hint

Check if the protocol has associated types or Self requirements.

The protocol Runnable has no associated types or Self requirements, so it can be used as a type without 'any' keyword in Swift 5.6 and earlier. In Swift 5.7+, 'any' is recommended but not mandatory here.

❓ Predict Output

advanced

Output when using 'any' with protocol with associated type

What happens when you try to compile and run this Swift code?

Swift

protocol Container {
    associatedtype Item
    func getItem() -> Item
}

struct Box: Container {
    func getItem() -> Int {
        42
    }
}

func printItem(container: any Container) {
    print(container.getItem())
}

let box = Box()
printItem(container: box)

A Compile-time error: Missing 'any' keyword before 'Container'

B Prints 42

C Runtime error: cannot call getItem() on existential

D Compile-time error: Protocol 'Container' can only be used as a generic constraint because it has Self or associated type requirements

💡 Hint

Protocols with associated types cannot be used as existential types directly.

Because Container has an associated type, it cannot be used as an existential type with 'any'. The compiler produces an error.

🧠 Conceptual

expert

Why use 'any' keyword for existential types in Swift?

Which of the following best explains the purpose of the 'any' keyword in Swift existential types?

A It explicitly marks a protocol type as an existential to improve code clarity and future-proofing

B It enables protocols with associated types to be used as types directly

C It automatically converts protocol types into generic types

D It disables protocol conformance checking at compile time

💡 Hint

Think about why Swift introduced the 'any' keyword in version 5.7.

The 'any' keyword makes it clear when a protocol is used as an existential type, improving readability and helping the compiler distinguish between generic constraints and existential usage.