Protocol as types in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a protocol named Drivable.

Swift

protocol [1] {
    func drive()
}

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

ADriver

BDrivable

CDriveable

DDrive

Attempts:

3 left

💡 Hint

Common Mistakes

Using a noun like 'Driver' instead of a protocol name.

Misspelling the protocol name.

The protocol name should be Drivable to describe something that can be driven.

2fill in blank

medium

Complete the code to make Car conform to the Drivable protocol.

Swift

struct Car: [1] {
    func drive() {
        print("Car is driving")
    }
}

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

ADrivable

BDriving

CDriver

DDrive

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong protocol name.

Forgetting the colon before the protocol name.

To conform to a protocol, the type must list the protocol name after a colon.

3fill in blank

hard

Fix the error in the code by completing the function parameter type with the protocol name.

Swift

func startDriving(vehicle: [1]) {
    vehicle.drive()
}

let myCar = Car()
startDriving(vehicle: myCar)

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

ADriveable

BCar

CVehicle

DDrivable

Attempts:

3 left

💡 Hint

Common Mistakes

Using the concrete type 'Car' instead of the protocol.

Misspelling the protocol name.

Using the protocol as a type allows any conforming type to be passed in.

4fill in blank

hard

Fill both blanks to create a variable of protocol type and call its method.

Swift

var vehicle: [1] = Car()
vehicle.[2]()

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

ADrivable

Bdrive

Cstart

DCar

Attempts:

3 left

💡 Hint

Common Mistakes

Using the concrete type as variable type instead of the protocol.

Calling a method not declared in the protocol.

The variable should be declared as the protocol type and call the protocol method.

5fill in blank

hard

Fill all three blanks to define a protocol, a conforming struct, and use the protocol as a type.

Swift

protocol [1] {
    func [2]()
}

struct Bike: [1] {
    func [2]() {
        print("Bike is moving")
    }
}

let myVehicle: [1] = Bike()
myVehicle.[2]()

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

AMovable

Bmove

Cdrive

DRunnable

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names for protocol and method in different places.

Choosing method names not matching the protocol.

The protocol is named Movable with method move. The struct conforms to Movable and implements move. The variable uses Movable type and calls move.

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