Protocol as types in Swift

Introduction

Protocols let us define a set of rules or features. Using a protocol as a type means we can work with any object that follows those rules, without caring about its exact kind.

When you want to write a function that can accept different kinds of objects, as long as they share some behavior.

When you want to store different objects in the same list or variable, but only care about the shared features.

When you want to hide the exact type of an object and only expose what it can do.

When you want to make your code flexible and easy to change later.

Syntax

Swift

protocol SomeProtocol {
    func doSomething()
}

func useProtocolType(item: SomeProtocol) {
    item.doSomething()
}

You use the protocol name as a type to accept any object that conforms to it.

This helps write flexible and reusable code.

Examples

Here, Greetable is a protocol. The function sayHello accepts any object that can greet.

Swift

protocol Greetable {
    func greet()
}

struct Person: Greetable {
    func greet() {
        print("Hello!")
    }
}

func sayHello(to someone: Greetable) {
    someone.greet()
}

We create an array of Drawable objects. Each can be drawn, even if they are different shapes.

Swift

protocol Drawable {
    func draw()
}

class Circle: Drawable {
    func draw() {
        print("Drawing a circle")
    }
}

let shapes: [Drawable] = [Circle(), Circle()]
for shape in shapes {
    shape.draw()
}

Sample Program

This program defines a Vehicle protocol. Both Car and Bike follow it. The function startJourney accepts any Vehicle and calls its drive method.

Swift

protocol Vehicle {
    func drive()
}

struct Car: Vehicle {
    func drive() {
        print("Car is driving")
    }
}

struct Bike: Vehicle {
    func drive() {
        print("Bike is driving")
    }
}

func startJourney(with vehicle: Vehicle) {
    vehicle.drive()
}

let myCar = Car()
let myBike = Bike()

startJourney(with: myCar)
startJourney(with: myBike)

Important Notes

Protocols as types let you write code that works with many kinds of objects.

Only the methods and properties in the protocol are accessible when using the protocol as a type.

If you need to access something specific to a concrete type, you must cast it.

Summary

Protocols define a set of features or rules.

Using a protocol as a type means any object following those rules can be used.

This makes your code flexible and easier to maintain.

Practice

1. What does it mean to use a protocol as a type in Swift?

easy

A. Any object that follows the protocol's rules can be used as that type.

B. Only classes can conform to the protocol.

C. Protocols can only be used to create new classes.

D. Protocols define stored properties for types.

Solution

1. Step 1: Understand what a protocol defines

   A protocol defines a set of rules or features that a type must implement.

2. Step 2: Using protocol as a type

   When a protocol is used as a type, any object that implements those rules can be assigned to that type.

3. Final Answer:

   Any object that follows the protocol's rules can be used as that type. -> Option A

4. Quick Check:

   Protocol as type means flexible object use = A [OK]

Hint: Protocol as type means any conforming object fits [OK]

Common Mistakes:

• Thinking protocols only apply to classes
• Believing protocols define stored properties
• Confusing protocol usage with inheritance

2. Which of the following is the correct way to declare a variable of a protocol type named Drawable?

easy

A. var shape: Drawable

B. var shape: Drawable()

C. var shape: Drawable = Drawable()

D. var shape: Drawable[]

Solution

1. Step 1: Understand protocol type declaration

   To declare a variable of a protocol type, use the protocol name without parentheses or brackets.

2. Step 2: Analyze options

   var shape: Drawable correctly declares a variable of type Drawable. Options A and B incorrectly try to instantiate a protocol, which is not allowed. var shape: Drawable[] uses invalid syntax.

3. Final Answer:

   var shape: Drawable -> Option A

4. Quick Check:

   Protocol type variable = var name: ProtocolName [OK]

Hint: Declare protocol type without parentheses or initialization [OK]

Common Mistakes:

• Trying to instantiate a protocol directly
• Using parentheses after protocol name
• Using array brackets incorrectly

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

protocol Greetable {
    func greet() -> String
}

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

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

medium

A. greet()

B. Compile error

C. Person

D. Hello

Solution

1. Step 1: Understand protocol conformance

   Person struct conforms to Greetable and implements greet() returning "Hello".

2. Step 2: Using protocol as type and calling method

   Variable greeter is of type Greetable but holds a Person instance, so calling greet() returns "Hello".

3. Final Answer:

   Hello -> Option D

4. Quick Check:

   Protocol method call returns implemented string = Hello [OK]

Hint: Protocol variable calls conforming type's method [OK]

Common Mistakes:

• Expecting protocol name or method name as output
• Thinking protocol cannot be used as variable type
• Confusing compile error with runtime behavior

4. Identify the error in this Swift code using protocol as a type:

protocol Vehicle {
    func start()
}

class Car: Vehicle {
    func start() {
        print("Car started")
    }
}

var myVehicle: Vehicle = Vehicle()
myVehicle.start()

medium

A. Variable myVehicle should be a class, not protocol

B. Missing method implementation in Car

C. Cannot instantiate protocol directly with Vehicle()

D. start() method should return a value

Solution

1. Step 1: Check protocol instantiation

   Protocols cannot be instantiated directly, so Vehicle() causes an error.

2. Step 2: Analyze variable assignment

   Variable myVehicle must hold an instance of a type conforming to Vehicle, like Car(), not Vehicle() itself.

3. Final Answer:

   Cannot instantiate protocol directly with Vehicle() -> Option C

4. Quick Check:

   Protocols can't be created with () = Error [OK]

Hint: Protocols can't be instantiated directly [OK]

Common Mistakes:

• Trying to create an instance of a protocol
• Confusing protocol with class instantiation
• Ignoring method return types

5. Given the protocol and classes below, which code correctly stores different types conforming to Animal in an array and calls sound() on each?

protocol Animal {
    func sound() -> String
}

class Dog: Animal {
    func sound() -> String { "Bark" }
}

class Cat: Animal {
    func sound() -> String { "Meow" }
}

hard

A. let animals: [Animal] = [Dog(), Cat()]
for animal in animals {
print(animal.sound)
}

B. let animals: [Animal] = [Dog(), Cat()]
for animal in animals {
print(animal.sound())
}

C. let animals: [Dog] = [Dog(), Cat()]
for animal in animals {
print(animal.sound())
}

D. let animals = [Dog(), Cat()]
for animal in animals {
print(animal.sound())
}

Solution

1. Step 1: Understand protocol array usage

   To store different types conforming to Animal, declare array as [Animal].

2. Step 2: Call method correctly

   Use animal.sound() with parentheses to call the method. let animals: [Animal] = [Dog(), Cat()] for animal in animals { print(animal.sound()) } does this correctly.

3. Final Answer:

   let animals: [Animal] = [Dog(), Cat()] for animal in animals { print(animal.sound()) } -> Option B

4. Quick Check:

   Protocol array with method calls = let animals: [Animal] = [Dog(), Cat()] for animal in animals { print(animal.sound()) } [OK]

Hint: Use protocol array type and call methods with () [OK]

Common Mistakes:

• Using concrete type array instead of protocol type
• Calling method without parentheses
• Mixing incompatible types in array