Existential types (any keyword) in Swift - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is an existential type in Swift?

An existential type is a type that can hold any value conforming to a specific protocol or set of protocols. It allows you to work with values of different types through a common interface.

beginner

What does the any keyword do in Swift?

The any keyword explicitly marks a type as an existential type, meaning it can hold any value conforming to the specified protocol. It clarifies that the value is not a specific type but any type that fits the protocol.

beginner

How do you declare a variable that can hold any type conforming to a protocol Drawable using the any keyword?

You declare it like this:
var shape: any Drawable
This means shape can hold any value that conforms to the Drawable protocol.

intermediate

Why did Swift introduce the any keyword for existential types?

Swift introduced any to make it clear when a type is existential, improving code readability and safety by distinguishing between concrete types and existential types.

intermediate

Can you use existential types with multiple protocols? How?

Yes, you can combine multiple protocols using the any keyword like this:
var value: any ProtocolA & ProtocolB
This means value can hold any type conforming to both ProtocolA and ProtocolB.

What does the any keyword in Swift indicate?

A A type alias

B A concrete type is used

C A generic type parameter

D A value can be any type conforming to a protocol

The any keyword marks a type as existential, meaning it can hold any value conforming to the specified protocol.

Which of the following is a correct way to declare an existential type variable for protocol Equatable?

A var item: Equatable

B var item: <Equatable>

C var item: any Equatable

D var item: Equatable & any

Using any Equatable explicitly declares an existential type variable.

Why is using any keyword recommended over omitting it for existential types?

A It improves code clarity and safety

B It makes code run faster

C It allows inheritance

D It disables protocol conformance

The any keyword makes it clear the type is existential, improving readability and safety.

Can you assign a value of a concrete type conforming to a protocol to a variable declared as any Protocol?

A No, only existential types are allowed

B Yes, because the variable accepts any conforming type

C Only if the concrete type is a class

D Only if the concrete type is a struct

Existential types can hold any value conforming to the protocol, including concrete types.

How do you declare a variable that conforms to both Codable and Hashable using existential types?

A var data: any Codable & Hashable

B var data: Codable & Hashable

C var data: any Codable, Hashable

D var data: Codable | Hashable

You combine protocols with & and prefix with any to declare an existential type conforming to multiple protocols.

Explain what existential types are in Swift and how the any keyword is used with them.

Think about how you can hold different types that share a protocol.

Describe the benefits of using the any keyword explicitly in your Swift code.

Why does Swift want you to write <code>any</code> now?

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