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]

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

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)

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

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)

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

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.

Practice

1. In Swift, what happens when you assign one class instance to another variable?

easy

A. The compiler throws an error because classes cannot be assigned.

B. A new copy of the object is created; changes do not affect each other.

C. Both variables refer to the same object; changes affect both.

D. The original object is deleted and replaced by the new variable.

Solution

1. Step 1: Understand class assignment behavior

   In Swift, classes are reference types, so assigning one instance to another variable makes both variables point to the same object.

2. Step 2: Effect of modifying through one variable

   Changing the object through one variable affects the object seen by the other variable because they share the same reference.

3. Final Answer:

   Both variables refer to the same object; changes affect both. -> Option C

4. Quick Check:

   Class assignment shares references = D [OK]

Hint: Classes share references; variables point to the same object [OK]

Common Mistakes:

• Thinking classes copy data on assignment
• Confusing classes with structs
• Assuming changes affect only one variable

2. Which of the following is the correct way to declare a class instance and assign it to another variable in Swift?

easy

A. let obj1 = MyClass(); let obj2 = obj1

B. var obj1 = MyClass() var obj2 = obj1

C. let obj1 = MyClass() let obj2: MyClass = obj1()

D. var obj1 = MyClass() let obj2 = new MyClass()

Solution

1. Step 1: Check correct syntax for class instance creation

   In Swift, you create a class instance with MyClass() and assign it to a variable using let or var with an equals sign.

2. Step 2: Verify assignment to another variable

   Assigning obj1 to obj2 is done simply by let obj2 = obj1; no parentheses or 'new' keyword are used.

3. Final Answer:

   let obj1 = MyClass(); let obj2 = obj1 -> Option A

4. Quick Check:

   Correct class instance assignment syntax = A [OK]

Hint: Use let/var with equals and no 'new' keyword in Swift [OK]

Common Mistakes:

• Using 'new' keyword like other languages
• Adding parentheses after variable names
• Missing semicolons but Swift allows them optionally

3. Consider this Swift code:

class Box {
  var value: Int
  init(value: Int) { self.value = value }
}

let box1 = Box(value: 10)
let box2 = box1
box2.value = 20
print(box1.value)

What is the output?

medium

A. 20

B. 10

C. Compilation error

D. Runtime error

Solution

1. Step 1: Understand reference sharing in class instances

   box1 and box2 both refer to the same Box object because classes are reference types.

2. Step 2: Effect of changing box2.value

   Changing box2.value to 20 changes the single shared object, so box1.value also reflects 20.

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Class reference sharing means box1.value = 20 [OK]

Hint: Changing one reference changes all pointing to same object [OK]

Common Mistakes:

• Assuming box1.value stays 10 after box2 changes it
• Confusing class with struct behavior
• Expecting a compile or runtime error

4. What is wrong with this Swift code snippet?

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

var p1 = Person(name: "Alice")
var p2 = p1
p2.name = "Bob"
print(p1.name)

medium

A. p2 should be declared with let, not var.

B. Nothing is wrong; output is "Bob".

C. Changing p2.name does not affect p1.name.

D. Class Person must be a struct to allow copying.

Solution

1. Step 1: Analyze reference sharing in the code

   p1 and p2 both point to the same Person instance because Person is a class.

2. Step 2: Effect of changing p2.name

   Changing p2.name to "Bob" changes the shared object, so p1.name also becomes "Bob".

3. Final Answer:

   Nothing is wrong; output is "Bob". -> Option B

4. Quick Check:

   Class reference sharing means p1.name = "Bob" [OK]

Hint: Class instances share references; changes reflect everywhere [OK]

Common Mistakes:

• Expecting p1.name to remain "Alice"
• Thinking var vs let affects reference sharing
• Believing class must be struct for copying

5. You want to create a copy of a class instance in Swift so that modifying the copy does not affect the original. Which approach correctly achieves this?

hard

A. Declare the class as a struct instead of a class.

B. Assign the instance to a new variable; this automatically copies it.

C. Use the 'clone' keyword to duplicate the instance.

D. Make the class conform to NSCopying and implement copy(with:), then use copy().

Solution

1. Step 1: Understand copying behavior of classes

   Classes in Swift do not copy on assignment; variables share references by default.

2. Step 2: Use NSCopying protocol for explicit copying

   To create a true copy, the class must conform to NSCopying and implement copy(with:) method, then call copy() to get a new instance.

3. Step 3: Evaluate other options

   Assigning to a new variable shares the reference. Declaring as a struct changes the type but is not always possible. The 'clone' keyword does not exist in Swift.

4. Final Answer:

   Make the class conform to NSCopying and implement copy(with:), then use copy(). -> Option D

5. Quick Check:

   Explicit copy via NSCopying = C [OK]

Hint: Use NSCopying protocol to make class copies [OK]

Common Mistakes:

• Assuming assignment copies the object
• Confusing structs with classes for copying
• Expecting a 'clone' keyword in Swift