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.

Practice

1. What does the any keyword indicate in Swift when used with a protocol type?

easy

A. It creates a new protocol that inherits from the original.

B. It marks the type as existential, allowing any value conforming to the protocol.

C. It restricts the type to only one specific conforming type.

D. It converts the protocol into a concrete class.

Solution

1. Step 1: Understand the role of any in Swift

   The any keyword is used to mark a type as existential, meaning it can hold any value that conforms to a given protocol.

2. Step 2: Compare with other options

   The incorrect options describe behaviors unrelated to any. It does not create new protocols, restrict to one type, or convert protocols to classes.

3. Final Answer:

   It marks the type as existential, allowing any value conforming to the protocol. -> Option B

4. Quick Check:

   any means existential type [OK]

Hint: Remember: any means 'any conforming type' [OK]

Common Mistakes:

• Thinking any creates a new protocol
• Confusing any with type restriction
• Assuming any converts protocols to classes

2. Which of the following is the correct syntax to declare a variable that can hold any value conforming to the protocol Drawable using existential types?

easy

A. var shape: Drawable

B. var shape: Drawable & any

C. var shape: any Drawable

D. var shape: Drawable<any>

Solution

1. Step 1: Recall the syntax for existential types

   In Swift, to declare a variable that can hold any value conforming to a protocol, you use any ProtocolName. So here, it is any Drawable.

2. Step 2: Analyze each option

   var shape: Drawable misses any, so it is not explicit existential. var shape: Drawable & any and var shape: Drawable<any> have incorrect syntax. Only var shape: any Drawable uses the correct syntax.

3. Final Answer:

   var shape: any Drawable -> Option C

4. Quick Check:

   Existential syntax = any Protocol [OK]

Hint: Use any before protocol name for existential types [OK]

Common Mistakes:

• Omitting any keyword
• Placing any after protocol name
• Using angle brackets incorrectly

3. What will be the output of the following Swift code?

protocol Greetable {
    func greet() -> String
}

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

let greeter: any Greetable = Person()
print(greeter.greet())

medium

A. Hello

B. Error: Protocol method not implemented

C. Runtime error

D. nil

Solution

1. Step 1: Understand the protocol and struct

   The protocol Greetable requires a greet() method returning a String. The struct Person implements this method returning "Hello".

2. Step 2: Analyze the existential usage and output

   The variable greeter is declared as any Greetable and assigned a Person instance. Calling greeter.greet() calls the implemented method and returns "Hello".

3. Final Answer:

   Hello -> Option A

4. Quick Check:

   Existential calls implemented method = "Hello" [OK]

Hint: Existential calls protocol methods on conforming instance [OK]

Common Mistakes:

• Expecting a compile or runtime error
• Thinking protocol methods are not accessible
• Assuming nil is returned

4. Identify the error in the following Swift code using existential types and fix it:

protocol Runnable {
    func run()
}

struct Dog: Runnable {
    func run() { print("Dog runs") }
}

var runner: Runnable = Dog()
runner.run()

medium

A. Missing any keyword before Runnable in variable declaration.

B. Dog does not conform to Runnable protocol.

C. run() method should return a value.

D. Variable runner should be a constant (let) instead of var.

Solution

1. Step 1: Check variable declaration syntax

   Since Swift 5.7, existential types require the any keyword explicitly. The declaration var runner: Runnable is invalid; it should be var runner: any Runnable.

2. Step 2: Verify other parts

   Dog conforms correctly, run() returns Void (allowed), and var is allowed. So only missing any is the error.

3. Final Answer:

   Missing any keyword before Runnable in variable declaration. -> Option A

4. Quick Check:

   Existential types need any keyword [OK]

Hint: Add any before protocol for existential variables [OK]

Common Mistakes:

• Omitting any keyword
• Thinking protocol conformance is missing
• Expecting run() to return a value

5. You want to write a function that accepts any value conforming to the Codable protocol and returns its JSON string representation. Which is the correct way to declare the function parameter using existential types?

hard

A. func encodeToJSON(value: Codable & any) -> String

B. func encodeToJSON(value: Codable) -> String

C. func encodeToJSON(value: T) -> String

D. func encodeToJSON(value: any Codable) -> String

Solution

1. Step 1: Understand existential vs generic parameters

   Using any Codable declares an existential type parameter that can accept any value conforming to Codable. This matches the question requirement.

2. Step 2: Analyze each option

   func encodeToJSON(value: Codable) -> String omits any, which is invalid syntax for existential types. func encodeToJSON(value: T) -> String uses generics, which is valid but not existential. func encodeToJSON(value: Codable & any) -> String has incorrect syntax. func encodeToJSON(value: any Codable) -> String is correct.

3. Final Answer:

   func encodeToJSON(value: any Codable) -> String -> Option D

4. Quick Check:

   Existential parameter syntax = any Protocol [OK]

Hint: Use any before protocol for existential function parameters [OK]

Common Mistakes:

• Using generics instead of existential types
• Omitting any keyword
• Incorrect syntax with & or angle brackets