Existential types (any keyword) in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a variable that can hold any type conforming to the protocol.

Swift

var value: [1] CustomStringConvertible

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

Aany

Bsome

Clet

Dvar

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'some' instead of 'any' which is for opaque types.

Using 'var' or 'let' inside the type declaration.

The 'any' keyword is used to declare an existential type that can hold any value conforming to the protocol.

2fill in blank

medium

Complete the function signature to accept any type conforming to Equatable.

Swift

func compare(_ a: [1] Equatable, _ b: [1] Equatable) -> Bool

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

Alet

Bsome

Cvar

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'some' which is for opaque types and not allowed in parameter positions.

Omitting the keyword and causing a compile error.

The 'any' keyword is used to accept any value conforming to the Equatable protocol as an existential type.

3fill in blank

hard

Fix the error in the code by adding the correct keyword for existential types.

Swift

let description: [1] CustomStringConvertible = "Hello"

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

Asome

Bvar

Cany

Dlet

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'some' which is for opaque types and incompatible here.

Omitting the keyword causing a compiler error.

The 'any' keyword is required to declare an existential type variable holding a value conforming to CustomStringConvertible.

4fill in blank

hard

Complete the code to create a dictionary with keys as Strings and values as any type conforming to CustomStringConvertible.

Swift

let dict: [String: [1] CustomStringConvertible] = ["key":  "value"]

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

Aany

Bsome

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'some' which is for opaque types and not allowed here.

Omitting 'any' from the type declaration causing a compiler error.

The dictionary value type uses 'any' to indicate an existential type conforming to CustomStringConvertible. No keyword is needed before the string literal.

5fill in blank

hard

Fill both blanks to define a function returning any type conforming to Equatable and assign a value inside.

Swift

func getValue() -> [1] Equatable {
    let val: [2] Equatable = 42
    return  val
}

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

Aany

Dsome

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'some' which is for opaque types and incompatible here.

Adding 'any' in the return statement which is not needed.

The function return type and variable use 'any' for existential types. The return statement returns the variable directly without extra keywords.

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