String type and immutability in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a string variable named greeting.

Kotlin

val greeting: String = [1]

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

A'Hello, Kotlin!'

BHello, Kotlin!

C"Hello, Kotlin!"

DString("Hello, Kotlin!")

Attempts:

3 left

💡 Hint

Common Mistakes

Using single quotes instead of double quotes for strings.

Not enclosing the string in quotes at all.

In Kotlin, string literals must be enclosed in double quotes. So "Hello, Kotlin!" is the correct way to assign a string value.

2fill in blank

medium

Complete the code to get the length of the string stored in name.

Kotlin

val length = name.[1]

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

Alength

Bsize

Ccount

Dlength()

Attempts:

3 left

💡 Hint

Common Mistakes

Using length() as if it were a method.

Using size or count which are not properties of String.

The length property returns the number of characters in a Kotlin string.

3fill in blank

hard

Fix the error in the code that tries to change a character in a string.

Kotlin

var text = "Kotlin"
text[1] = 'C'

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

A{0}

B[0]

C(0)

D<0>

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to assign a character to a string index (strings are immutable).

Using parentheses or braces instead of square brackets.

Strings in Kotlin are immutable, so you cannot change characters directly. But if you want to access a character, you use square brackets [index]. However, assignment to a character at an index is not allowed.

4fill in blank

hard

Fill both blanks to create a new string by replacing the first character of original with 'C'.

Kotlin

val original = "Kotlin"
val modified = [1] + original.[2](1)

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

A"C"

Bsubstring

CsubSequence

DremoveRange

Attempts:

3 left

💡 Hint

Common Mistakes

Using subSequence instead of substring which returns CharSequence, not String.

Trying to modify the original string directly.

To replace the first character, we concatenate "C" with the substring of original starting from index 1.

5fill in blank

hard

Fill all three blanks to create a new string with all letters uppercase and check if it equals "KOTLIN".

Kotlin

val word = "kotlin"
val upperWord = word.[1]()
val isEqual = upperWord [2] "KOTLIN"
println(isEqual)  // prints [3]

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

Auppercase

B==

Ctrue

Dequals

Attempts:

3 left

💡 Hint

Common Mistakes

Using equals() method without parentheses.

Using equals instead of == for comparison.

Expecting the print output to be false.

The uppercase() function converts the string to uppercase. Then we compare with == to check equality. The result is true because they match.

Practice

1. What does it mean that a String in Kotlin is immutable?

easy

A. Strings can only hold numbers, not letters.

B. You can change characters inside the String directly.

C. Once created, the text inside a String cannot be changed.

D. Strings automatically update when variables change.

Solution

1. Step 1: Understand immutability concept

   Immutability means the object cannot be changed after creation.

2. Step 2: Apply to Kotlin String

   Kotlin Strings cannot be modified once created; any change creates a new String.

3. Final Answer:

   Once created, the text inside a String cannot be changed. -> Option C

4. Quick Check:

   String immutability = cannot change text [OK]

Hint: Remember: Strings are like printed paper, not editable [OK]

Common Mistakes:

• Thinking you can change characters inside the same String
• Confusing immutability with variable reassignment
• Believing Strings hold only numbers

2. Which of the following is the correct way to concatenate two Strings str1 and str2 in Kotlin?

easy

A. str1 & str2

B. str1 + str2

C. str1.join(str2)

D. str1.append(str2)

Solution

1. Step 1: Recall Kotlin String concatenation syntax

   Kotlin uses the + operator to join two Strings.

2. Step 2: Check each option

   Only str1 + str2 is valid Kotlin syntax for concatenation.

3. Final Answer:

   str1 + str2 -> Option B

4. Quick Check:

   Use + to join Strings [OK]

Hint: Use + to combine Strings, like adding words together [OK]

Common Mistakes:

• Using methods like append or join which don't exist on Kotlin String
• Using & which is not a Kotlin operator for Strings
• Trying to modify Strings directly

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

val text = "Hello"
val newText = text + " World"
println(text)

medium

A. Hello

B. Hello World

C. World

D. HelloWorld

Solution

1. Step 1: Understand String immutability and concatenation

   The original String text remains unchanged after concatenation.

2. Step 2: Analyze the print statement

   Only text is printed, which is still "Hello".

3. Final Answer:

   Hello -> Option A

4. Quick Check:

   Original String unchanged = Hello [OK]

Hint: Printing original String shows it stays the same [OK]

Common Mistakes:

• Assuming original String changes after + operation
• Confusing newText with text in print
• Ignoring immutability of Strings

4. Identify the error in this Kotlin code snippet:

var greeting = "Hi"
greeting[0] = 'B'
println(greeting)

medium

A. println cannot print Strings.

B. Variable greeting should be val, not var.

C. Single quotes cannot be used for characters.

D. You cannot change characters in a String by index.

Solution

1. Step 1: Check String mutability rules

   Kotlin Strings are immutable; you cannot assign to an index.

2. Step 2: Analyze the code line causing error

   Line greeting[0] = 'B' tries to change a character, which is invalid.

3. Final Answer:

   You cannot change characters in a String by index. -> Option D

4. Quick Check:

   Strings immutable = no index assignment [OK]

Hint: Strings are like frozen text, can't change letters inside [OK]

Common Mistakes:

• Thinking var allows character change inside String
• Confusing char literals with strings
• Believing println can't print Strings

5. Given a list of words, how can you create a new list with each word appended by "!" without changing the original words?

val words = listOf("cat", "dog", "bird")
val excitedWords = ???

Which Kotlin code correctly does this considering String immutability?

hard

A. val excitedWords = words.map { it + "!" }

B. val excitedWords = words.apply { this[0] += "!" }

C. val excitedWords = words.forEach { it + "!" }

D. val excitedWords = words.replace("", "!")

Solution

1. Step 1: Understand immutability and list transformations

   Strings cannot be changed, so create new Strings with + "!".

2. Step 2: Use map to transform each word into a new String

   map returns a new list with transformed elements without modifying original.

3. Final Answer:

   val excitedWords = words.map { it + "!" } -> Option A

4. Quick Check:

   Use map + + operator for new list [OK]

Hint: Use map to create new list with modified Strings [OK]

Common Mistakes:

• Trying to modify list elements directly with apply or forEach
• Using replace on list instead of Strings
• Assuming original list changes