Codable protocol for encoding/decoding in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to make the struct conform to Codable.

Swift

struct User: [1] {
    var name: String
    var age: Int
}

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

ACodable

BEquatable

CHashable

DIdentifiable

Attempts:

3 left

💡 Hint

Common Mistakes

Using Equatable instead of Codable.

Forgetting to conform to any protocol.

The Codable protocol allows the struct to be encoded and decoded easily.

2fill in blank

medium

Complete the code to encode the user instance to JSON data.

Swift

let user = User(name: "Alice", age: 30)
let encoder = JSONEncoder()
let data = try? encoder.[1](user)

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

Aencode

Bserialize

Cdecode

Dconvert

Attempts:

3 left

💡 Hint

Common Mistakes

Using decode instead of encode.

Trying to use serialize or convert which are not methods of JSONEncoder.

The encode method converts the instance into JSON data.

3fill in blank

hard

Fix the error in decoding JSON data to a User instance.

Swift

let decoder = JSONDecoder()
let user = try? decoder.[1](User.self, from: jsonData)

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

Aencode

Bserialize

Cdecode

Dconvert

Attempts:

3 left

💡 Hint

Common Mistakes

Using encode instead of decode.

Trying to use serialize or convert which are not methods of JSONDecoder.

The decode method is used to convert JSON data back into a Swift instance.

4fill in blank

hard

Fill both blanks to create a dictionary from an array of users with their names as keys.

Swift

let users = [User(name: "Bob", age: 25), User(name: "Carol", age: 28)]
let userDict = Dictionary(uniqueKeysWithValues: users.map { [1] in ([1].[2], [1].age) })

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Auser

Bname

Cage

Ditem

Attempts:

3 left

💡 Hint

Common Mistakes

Using property names instead of variable names in the closure parameters.

Mixing up the order of variables.

We use user and name as names in the closure to map users to a dictionary with names as keys and ages as values.

5fill in blank

hard

Fill all three blanks to encode a User instance and convert it to a JSON string.

Swift

let user = User(name: "Dave", age: 40)
let encoder = JSONEncoder()
if let data = try? encoder.[1](user) {
    let jsonString = String(data: data, encoding: [2])
    print(jsonString ?? [3])
}

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

Aencode

Butf8

C"Invalid JSON"

Ddecode

Attempts:

3 left

💡 Hint

Common Mistakes

Using decode instead of encode.

Using wrong encoding like ascii.

Not providing a fallback string.

We encode the user, convert data to a UTF-8 string, and print a fallback message if conversion fails.