POP vs OOP decision patterns in Swift - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a protocol named 'Vehicle'.

Swift

protocol [1] {
    func drive()
}

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

AVehicle

BCar

CDriveable

DTransport

Attempts:

3 left

💡 Hint

Common Mistakes

Using a class name instead of a protocol name

Naming the protocol something unrelated

The protocol is named 'Vehicle' to define a common interface for types that can drive.

2fill in blank

medium

Complete the code to make 'Car' conform to the 'Vehicle' protocol.

Swift

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

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

ADriveable

BTransport

CVehicle

DMovable

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different protocol name

Forgetting to conform to the protocol

'Car' conforms to the 'Vehicle' protocol by adopting its name after the colon.

3fill in blank

hard

Fix the error in the class inheritance to use OOP correctly.

Swift

class [1] {
    func drive() {
        print("Driving")
    }
}

class SportsCar: Vehicle {
}

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

ACar

BVehicle

CDriveable

DTransport

Attempts:

3 left

💡 Hint

Common Mistakes

Naming the base class differently than the inherited class

Using a protocol name instead of a class name

The base class should be named 'Vehicle' so that 'SportsCar' can inherit from it.

4fill in blank

hard

Fill both blanks to create a protocol extension that provides a default implementation.

Swift

protocol Vehicle {
    func drive()
}

extension Vehicle {
    func [1]() {
        print("[2] driving")
    }
}

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

Adrive

Bstart

CVehicle

DCar

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different method name

Printing the wrong type name

The extension provides a default 'drive' method printing 'Vehicle driving'.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension filtering vehicles by speed.

Swift

let vehicles = ["car": 120, "bike": 80, "truck": 100]
let fastVehicles = vehicles.filter { $0.value [1] [2] }.map { $0.key.uppercased() }
let result = fastVehicles.reduce(into: [[3]: true]) { $0[$1] = true }

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

A>

B90

C""

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>'

Using a wrong threshold value

Using wrong dictionary key type

The code filters vehicles with speed greater than 90, converts keys to uppercase, and creates a dictionary with keys as uppercase vehicle names.