Existential types (any keyword) in Swift - Mini Project: Build & Apply

Using Existential Types with the any Keyword in Swift

📖 Scenario: Imagine you are building a simple app that manages different types of shapes. Each shape can tell its area, but the shapes are different types. You want to store them together and work with them in a common way.

🎯 Goal: You will create a protocol for shapes, then use the any keyword to hold different shapes in one list. Finally, you will calculate and print the area of each shape.

📋 What You'll Learn

Create a protocol called Shape with a property area of type Double

Create two structs Circle and Rectangle that conform to Shape

Create a variable shapes that holds an array of any Shape

Use a for loop to print the area of each shape

💡 Why This Matters

🌍 Real World

Using existential types with the <code>any</code> keyword helps you write flexible code that can work with different types sharing common behavior, like shapes in a drawing app.

💼 Career

Understanding existential types is important for Swift developers to handle collections of different types safely and clearly, which is common in app development and frameworks.

1

Create the Shape protocol and two shape structs

Create a protocol called Shape with a read-only property area of type Double. Then create two structs called Circle and Rectangle that conform to Shape. Circle should have a property radius of type Double, and Rectangle should have properties width and height of type Double. Implement the area property for both structs to calculate the correct area.

Swift

# Your code here to create protocol Shape and structs Circle and Rectangle

Hint

Remember, a protocol defines a property called area. The structs must calculate area using the formulas for circle and rectangle.

2

Create an array of any Shape

Create a variable called shapes that holds an array of any Shape. Initialize it with a Circle of radius 3.0 and a Rectangle of width 4.0 and height 5.0.

Swift

protocol Shape {
    var area: Double { get }
}

struct Circle: Shape {
    var radius: Double
    var area: Double {
        return Double.pi * radius * radius
    }
}

struct Rectangle: Shape {
    var width: Double
    var height: Double
    var area: Double {
        return width * height
    }
}

// Create the shapes array here
# Your code here

Hint

Use the syntax [any Shape] to hold different shapes in one array.

3

Loop over shapes and print their areas

Use a for loop with the variable shape to iterate over shapes. Inside the loop, print the area of each shape using print("Area: \(shape.area)").

Swift

protocol Shape {
    var area: Double { get }
}

struct Circle: Shape {
    var radius: Double
    var area: Double {
        return Double.pi * radius * radius
    }
}

struct Rectangle: Shape {
    var width: Double
    var height: Double
    var area: Double {
        return width * height
    }
}

var shapes: [any Shape] = [Circle(radius: 3.0), Rectangle(width: 4.0, height: 5.0)]

// Loop over shapes and print area
# Your code here

Hint

Use a for loop to go through each shape and print its area.

4

Print the areas of all shapes

Run the program to print the area of each shape in the shapes array. The output should show the area of the circle and the rectangle.

Swift

protocol Shape {
    var area: Double { get }
}

struct Circle: Shape {
    var radius: Double
    var area: Double {
        return Double.pi * radius * radius
    }
}

struct Rectangle: Shape {
    var width: Double
    var height: Double
    var area: Double {
        return width * height
    }
}

var shapes: [any Shape] = [Circle(radius: 3.0), Rectangle(width: 4.0, height: 5.0)]

for shape in shapes {
    print("Area: \(shape.area)")
}

// Run the program and check output

Hint

The program should print the area of the circle (about 28.27) and the rectangle (20.0).

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