It keyword for single parameter in Kotlin - Mini Project: Build & Apply

Using the It Keyword for Single Parameter in Kotlin

📖 Scenario: You are working on a simple Kotlin program that processes a list of names. You want to transform each name to uppercase using a concise way to handle single parameters in lambda expressions.

🎯 Goal: Learn how to use the it keyword in Kotlin lambda expressions when there is only one parameter.

📋 What You'll Learn

Create a list of names with exact values

Create a lambda expression using the it keyword to convert names to uppercase

Store the transformed names in a new list

Print the new list

💡 Why This Matters

🌍 Real World

Using the <code>it</code> keyword helps write shorter and clearer code when working with collections, such as lists of names, products, or messages.

💼 Career

Many Kotlin jobs require working with collections and lambdas. Knowing how to use <code>it</code> makes your code concise and easier to read, which is valued in professional Kotlin development.

1

Create a list of names

Create a list called names with these exact values: "alice", "bob", "charlie"

Kotlin

# Create a list called names with "alice", "bob", and "charlie"
# Your code here

Hint

Use listOf to create a list with the given strings.

2

Create a lambda expression using the it keyword

Create a new list called upperNames by calling names.map and use a lambda expression with the it keyword to convert each name to uppercase using it.uppercase()

Kotlin

val names = listOf("alice", "bob", "charlie")
# Create upperNames by mapping names with it.uppercase()
# Your code here

Hint

Use map with a lambda that uses it to refer to each element.

3

Print the transformed list

Print the upperNames list using println(upperNames)

Kotlin

val names = listOf("alice", "bob", "charlie")
val upperNames = names.map { it.uppercase() }
# Print the upperNames list
# Your code here

Hint

Use println to display the list.

4

Use the it keyword in a filter lambda

Create a new list called filteredNames by calling names.filter and use a lambda with the it keyword to keep only names that start with the letter 'a' using it.startsWith('a'). Then print filteredNames using println(filteredNames)

Kotlin

val names = listOf("alice", "bob", "charlie")
val upperNames = names.map { it.uppercase() }
println(upperNames)
# Create filteredNames by filtering names with it.startsWith('a')
# Print filteredNames
# Your code here

Hint

Use filter with a lambda using it to check the first letter.

Practice

1. What does the it keyword represent in a Kotlin lambda with a single parameter?

easy

A. The single implicit parameter of the lambda

B. A reserved word for loop counters

C. A keyword to declare variables

D. A function to print output

Solution

1. Step 1: Understand lambda parameters

   In Kotlin, if a lambda has only one parameter, you don't need to name it explicitly.

2. Step 2: Role of it

   The keyword it automatically refers to that single parameter inside the lambda.

3. Final Answer:

   The single implicit parameter of the lambda -> Option A

4. Quick Check:

   it = single lambda parameter [OK]

Hint: Remember: it means the one lambda input [OK]

Common Mistakes:

• Thinking it is a loop variable
• Confusing it with a keyword to declare variables
• Assuming it prints output

2. Which of the following is the correct syntax to use it in a Kotlin lambda that doubles each number in a list?

easy

A. list.map { it * 2 }

B. list.map { x -> x * 2 }

C. list.map { it -> it * 2 }

D. list.map { * 2 }

Solution

1. Step 1: Identify lambda with single parameter

   The lambda inside map has one parameter, so it can be used implicitly.

2. Step 2: Check syntax correctness

   list.map { it * 2 } correctly uses it without naming the parameter explicitly.

3. Final Answer:

   list.map { it * 2 } -> Option A

4. Quick Check:

   Implicit single parameter lambda syntax = list.map { it * 2 } [OK]

Hint: Use it only when lambda has one parameter [OK]

Common Mistakes:

• Using explicit parameter name and it together
• Omitting parameter name but adding arrow
• Trying to multiply without parameter

3. What is the output of this Kotlin code?

val numbers = listOf(1, 2, 3)
val result = numbers.filter { it > 1 }
println(result)

medium

A. [1, 2, 3]

B. [1]

C. [2, 3]

D. []

Solution

1. Step 1: Understand the filter condition

   The lambda filters numbers where it > 1, so only numbers greater than 1 remain.

2. Step 2: Apply filter to the list

   From the list [1, 2, 3], numbers 2 and 3 satisfy the condition.

3. Final Answer:

   [2, 3] -> Option C

4. Quick Check:

   Filter with it > 1 = [2, 3] [OK]

Hint: Filter keeps items where it meets condition [OK]

Common Mistakes:

• Including 1 in the result
• Confusing filter with map
• Expecting original list unchanged

4. Identify the error in this Kotlin code snippet:

val list = listOf(1, 2, 3)
val doubled = list.map { it * }
println(doubled)

medium

A. Missing parameter name before *

B. No error, code runs fine

C. Using it without declaring lambda parameter

D. Syntax error: incomplete expression after *

Solution

1. Step 1: Analyze the lambda expression

   The expression it * is incomplete; it lacks a value after the multiplication operator.

2. Step 2: Identify syntax error

   This causes a syntax error because Kotlin expects a complete expression after *.

3. Final Answer:

   Syntax error: incomplete expression after * -> Option D

4. Quick Check:

   Incomplete expression causes syntax error [OK]

Hint: Check for complete expressions after operators [OK]

Common Mistakes:

• Assuming it * is valid alone
• Thinking it must be declared explicitly
• Ignoring syntax errors

5. Given a list of strings val words = listOf("apple", "banana", "cat"), which Kotlin code uses it correctly to create a list of word lengths greater than 3?

hard

A. words.map { it.length > 3 }

B. words.filter { it.length > 3 }

C. words.map { it.length }.filter { it > 3 }

D. words.filter { length > 3 }.map { it }

Solution

1. Step 1: Understand the goal

   We want words whose lengths are greater than 3, so we filter words based on their length.

2. Step 2: Analyze each option

   words.filter { it.length > 3 }, words.filter { it.length > 3 }, correctly filters the words list to include only words with length greater than 3, using it properly.

3. Final Answer:

   words.filter { it.length > 3 } -> Option B

4. Quick Check:

   Filter words by length > 3 = words.filter { it.length > 3 } [OK]

Hint: Use filter with it.length > 3 to get words longer than 3 [OK]

Common Mistakes:

• Filtering on string directly without length
• Using it incorrectly in filter
• Mixing order of map and filter