Reference sharing and side effects in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a class instance and assign it to a variable.

Swift

class Person {
    var name: String
    init(name: String) {
        self.name = name
    }
}

let person1 = [1](name: "Alice")

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

AString

BArray

CInt

DPerson

Attempts:

3 left

💡 Hint

Common Mistakes

Using a type like String or Int instead of the class name.

Forgetting to call the initializer with parentheses.

We create an instance of the Person class by calling Person(name: "Alice").

2fill in blank

medium

Complete the code to assign one class instance to another variable, sharing the reference.

Swift

let person2 = person1
person2.name = "Bob"
print(person1.name)  // Output: [1]

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A"Unknown"

B"Bob"

C"Charlie"

D"Alice"

Attempts:

3 left

💡 Hint

Common Mistakes

Thinking that assigning creates a copy of the instance.

Expecting the original name to stay "Alice".

Both person1 and person2 refer to the same instance, so changing person2.name changes person1.name.

3fill in blank

hard

Fix the error in the code to prevent side effects by copying the instance.

Swift

class Person {
    var name: String
    init(name: String) {
        self.name = name
    }
    func copy() -> Person {
        return [1](name: self.name)
    }
}

let person1 = Person(name: "Alice")
let person2 = person1.copy()
person2.name = "Bob"
print(person1.name)  // Output: "Alice"

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

APerson

Bself

Ccopy

Dname

Attempts:

3 left

💡 Hint

Common Mistakes

Returning self instead of a new instance.

Calling copy() recursively inside itself.

The copy() method creates a new Person instance with the same name, avoiding shared references.

4fill in blank

hard

Fill in the blank to populate the dictionary with names only if they start with 'A'.

Swift

let names = ["Alice", "Bob", "Anna", "Charlie"]
var filteredNames = [String: String]()
for name in names {
  if name.[1]("A") {
    filteredNames[name] = name
  }
}
print(filteredNames)

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Acontains

BstartsWith

ChasPrefix

DendsWith

Attempts:

3 left

💡 Hint

Common Mistakes

Using non-existent methods like startsWith.

Using contains which checks anywhere in the string.

In Swift, hasPrefix("A") checks if a string starts with "A".

5fill in blank

hard

Fill all three blanks to create a dictionary of name lengths for names longer than 3 characters.

Swift

let names = ["Eve", "Alice", "Bob", "Charlie"]
var nameLengths = [String: Int]()
for [3] in names {
  if [2] > 3 {
    nameLengths[[1]] = [2]
  }
}
print(nameLengths)

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Aname

Bname.count

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using undefined variables like length.

Mixing up keys and values in the dictionary.

The loop uses name as the loop variable and key, name.count as the condition and value.