List creation (listOf, mutableListOf) in Kotlin - Mini Project: Build & Apply

List creation (listOf, mutableListOf)

📖 Scenario: You are organizing a small party and need to keep track of the guests who will attend.

🎯 Goal: Create a list of guests using listOf and then create a mutable list using mutableListOf to add more guests later.

📋 What You'll Learn

Create an immutable list called guests with exactly these names: "Alice", "Bob", "Charlie"

Create a mutable list called moreGuests starting with the same names as guests

Add the name "Diana" to the moreGuests list

Print the moreGuests list to show all guests

💡 Why This Matters

🌍 Real World

Lists are used everywhere to keep track of groups of items, like guests at a party, tasks to do, or products in a store.

💼 Career

Knowing how to create and modify lists is a basic skill for any programmer, useful in almost all software development jobs.

1

Create an immutable list of guests

Create an immutable list called guests using listOf with these exact names: "Alice", "Bob", "Charlie".

Kotlin

# Create the immutable list guests with the names Alice, Bob, and Charlie
# Your code here

Hint

Use val guests = listOf("Alice", "Bob", "Charlie") to create the list.

2

Create a mutable list from the immutable list

Create a mutable list called moreGuests using mutableListOf and initialize it with the contents of the guests list.

Kotlin

val guests = listOf("Alice", "Bob", "Charlie")
# Create a mutable list moreGuests initialized with guests
# Your code here

Hint

Use mutableListOf().apply { addAll(guests) } to copy the guests list.

3

Add a new guest to the mutable list

Add the name "Diana" to the moreGuests mutable list using the add method.

Kotlin

val guests = listOf("Alice", "Bob", "Charlie")
val moreGuests = mutableListOf().apply { addAll(guests) }
# Add "Diana" to moreGuests
# Your code here

Hint

Use moreGuests.add("Diana") to add the new guest.

4

Print the final list of guests

Print the moreGuests list to show all the guests including the newly added one.

Kotlin

val guests = listOf("Alice", "Bob", "Charlie")
val moreGuests = mutableListOf().apply { addAll(guests) }
moreGuests.add("Diana")
# Print the moreGuests list
# Your code here

Hint

Use println(moreGuests) to print the list.

Practice

1. Which Kotlin function creates a list that cannot be changed after it is created?

easy

A. listOf

B. mutableListOf

C. arrayListOf

D. setOf

Solution

1. Step 1: Understand listOf behavior

   listOf creates a fixed list that cannot be changed after creation.

2. Step 2: Compare with mutableListOf

   mutableListOf creates a list that can be changed (add/remove items).

3. Final Answer:

   listOf -> Option A

4. Quick Check:

   Fixed list = listOf [OK]

Hint: Use listOf for fixed lists, mutableListOf for changeable lists [OK]

Common Mistakes:

• Confusing mutableListOf as fixed list
• Thinking arrayListOf is immutable
• Using setOf which creates a set, not a list

2. Which of the following is the correct syntax to create a mutable list of integers with values 1, 2, and 3?

easy

A. val list = listOf(1, 2, 3)

B. val list = mutableListOf(1, 2, 3)

C. val list = mutableList(1, 2, 3)

D. val list = list(1, 2, 3)

Solution

1. Step 1: Identify correct function name

   The correct function to create a mutable list is mutableListOf.

2. Step 2: Check syntax correctness

   Options C and D use invalid function names. val list = listOf(1, 2, 3) creates an immutable list.

3. Final Answer:

   val list = mutableListOf(1, 2, 3) -> Option B

4. Quick Check:

   Mutable list syntax = mutableListOf(...) [OK]

Hint: Remember mutableListOf has 'Of' at the end [OK]

Common Mistakes:

• Using listOf instead of mutableListOf for mutable lists
• Misspelling mutableListOf as mutableList
• Using list() which does not exist

3. What will be the output of this Kotlin code?

val list = listOf("apple", "banana", "cherry")
println(list[1])

medium

A. banana

B. IndexOutOfBoundsException

C. cherry

D. apple

Solution

1. Step 1: Understand list indexing

   Kotlin lists use zero-based indexing, so index 1 refers to the second item.

2. Step 2: Identify the item at index 1

   The list is ["apple", "banana", "cherry"], so index 1 is "banana".

3. Final Answer:

   banana -> Option A

4. Quick Check:

   Index 1 = second item = banana [OK]

Hint: Remember list indexes start at 0 in Kotlin [OK]

Common Mistakes:

• Thinking index 1 is the first item
• Expecting an error for valid index
• Confusing listOf with mutableListOf output

4. What is wrong with this Kotlin code?

val list = listOf(1, 2, 3)
list.add(4)

medium

A. add() method requires a mutableListOf list

B. Syntax error in listOf declaration

C. listOf creates an immutable list; add() is not allowed

D. No error; code runs fine

Solution

1. Step 1: Check list type

   listOf creates an immutable (read-only) list.

2. Step 2: Understand add() method

   The add() method is only available on mutable lists, not on immutable lists.

3. Final Answer:

   listOf creates an immutable list; add() is not allowed -> Option C

4. Quick Check:

   Immutable list cannot add items [OK]

Hint: Use mutableListOf to add or remove items [OK]

Common Mistakes:

• Trying to add to listOf list
• Ignoring immutability of listOf
• Assuming add() works on all lists

5. You want to create a list of strings that you can add items to later. Which Kotlin code correctly creates this list and adds "orange" to it?

hard

A. val fruits = mutableListOf("apple", "banana")
fruits = fruits + "orange"

B. val fruits = listOf("apple", "banana")
fruits.add("orange")

C. val fruits = listOf("apple", "banana")
fruits += "orange"

D. val fruits = mutableListOf("apple", "banana")
fruits.add("orange")

Solution

1. Step 1: Choose a mutable list

   To add items later, the list must be mutable, so use mutableListOf.

2. Step 2: Use add() method correctly

   Calling add("orange") on a mutable list adds the item successfully.

3. Step 3: Check other options

   val fruits = listOf("apple", "banana") fruits.add("orange") tries to add to immutable list (error). val fruits = listOf("apple", "banana") fruits += "orange" uses += on immutable list (creates new list but does not modify original). val fruits = mutableListOf("apple", "banana") fruits = fruits + "orange" tries to reassign val which is not allowed.

4. Final Answer:

   val fruits = mutableListOf("apple", "banana") fruits.add("orange") -> Option D

5. Quick Check:

   Mutable list + add() = correct way [OK]

Hint: Use mutableListOf and add() to change list after creation [OK]

Common Mistakes:

• Adding to immutable list
• Trying to reassign val variable
• Using += on immutable list expecting mutation