Type checking with is operator in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Kotlin code using 'is' operator?

Consider the following Kotlin code snippet. What will it print?

Kotlin

fun main() {
    val obj: Any = "Hello"
    if (obj is String) {
        println(obj.length)
    } else {
        println("Not a string")
    }
}

A 5

B Not a string

C Compilation error

D Runtime exception

💡 Hint

Remember that 'is' checks if the object is of a certain type and smart casts it.

The variable 'obj' is of type Any but holds a String. The 'is String' check succeeds, so inside the if block, 'obj' is smart cast to String and 'obj.length' prints 5.

❓ Predict Output

intermediate

What does this Kotlin code print when using 'is' with nullable type?

Analyze the output of this Kotlin code snippet:

Kotlin

fun main() {
    val obj: Any? = null
    if (obj is String) {
        println("String with length ${obj.length}")
    } else {
        println("Not a string or null")
    }
}

A NullPointerException at runtime

B String with length 0

C Compilation error due to null check

D Not a string or null

💡 Hint

Check how 'is' operator behaves with null values.

The variable 'obj' is null. The 'is String' check returns false for null, so the else block runs printing 'Not a string or null'.

🔧 Debug

advanced

Why does this Kotlin code cause a compilation error?

Examine the code below and identify why it does not compile:

Kotlin

fun main() {
    val obj: Any = 123
    if (obj is String) {
        println(obj.length)
    } else {
        println(obj.length)
    }
}

A The variable 'obj' is immutable and cannot be checked

B Smart cast does not work inside if blocks

C The else block tries to access 'length' on 'Any' which is not guaranteed to have it

D The 'is' operator cannot be used with 'Any' type

💡 Hint

Consider what type 'obj' has in the else block and if 'length' is accessible.

In the else block, 'obj' is still of type 'Any' which does not have a 'length' property, causing a compilation error.

🧠 Conceptual

advanced

What is the behavior of 'is' operator with inheritance in Kotlin?

Given the classes below, what will the output be?

Kotlin

open class Animal
class Dog : Animal()

fun main() {
    val pet: Animal = Dog()
    if (pet is Dog) {
        println("It's a dog")
    } else {
        println("It's not a dog")
    }
}

A It's a dog

B It's not a dog

C Compilation error due to inheritance

D Runtime exception

💡 Hint

The 'is' operator checks the actual object type, not the declared type.

The variable 'pet' is declared as Animal but holds a Dog instance. The 'is Dog' check returns true, so it prints 'It's a dog'.

📝 Syntax

expert

Which option causes a syntax error in Kotlin when using 'is' operator?

Identify which code snippet below will cause a syntax error:

A if (obj is String) println(obj.length)

B if obj is String { println(obj.length) }

C if (obj is String) { println(obj.length) }

D if (obj is String) println(obj.length) else println("Not a string")

💡 Hint

Check the syntax of the if statement in Kotlin.

Option B misses parentheses around the condition, which is required in Kotlin if statements, causing a syntax error.

Practice

1. What does the is operator do in Kotlin?

easy

A. Checks if a variable is of a specific type

B. Changes the type of a variable

C. Creates a new variable

D. Deletes a variable

Solution

1. Step 1: Understand the purpose of is operator

   The is operator is used to check if a variable holds a value of a certain type.

2. Step 2: Differentiate from other options

   It does not change, create, or delete variables; it only checks type.

3. Final Answer:

   Checks if a variable is of a specific type -> Option A

4. Quick Check:

   Type check = A [OK]

Hint: Remember: is means 'is this type?' check [OK]

Common Mistakes:

• Thinking is changes the variable type
• Confusing is with variable creation
• Assuming is deletes variables

2. Which of the following is the correct syntax to check if obj is a String in Kotlin?

easy

A. if (obj instanceof String)

B. if (obj is String)

C. if (obj type String)

D. if (obj == String)

Solution

1. Step 1: Recall Kotlin syntax for type checking

   Kotlin uses is keyword to check types, not instanceof or other keywords.

2. Step 2: Match the correct syntax

   The correct syntax is if (obj is String).

3. Final Answer:

   if (obj is String) -> Option B

4. Quick Check:

   Correct syntax = C [OK]

Hint: Use is keyword, not Java's instanceof [OK]

Common Mistakes:

• Using Java syntax instanceof in Kotlin
• Writing obj type String which is invalid
• Comparing with == instead of type check

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

fun checkType(x: Any) {
    if (x is Int) {
        println(x + 5)
    } else {
        println("Not an Int")
    }
}

checkType(10)
checkType("Hello")

medium

A. Error at runtime

B. 10
Not an Int

C. 15
Hello5

D. 15
Not an Int

Solution

1. Step 1: Analyze the first call checkType(10)

   10 is an Int, so inside the if block, x + 5 equals 15, printed as "15".

2. Step 2: Analyze the second call checkType("Hello")

   "Hello" is not an Int, so else block runs printing "Not an Int".

3. Final Answer:

   15 Not an Int -> Option D

4. Quick Check:

   Int check prints sum, else prints message = A [OK]

Hint: If is true, use variable as that type inside block [OK]

Common Mistakes:

• Assuming string "Hello" can be added to 5
• Expecting runtime error instead of else output
• Confusing output order

4. Identify the error in this Kotlin code snippet:

fun printLength(obj: Any) {
    if (obj is String) {
        println(obj.length())
    }
}

medium

A. No error, code is correct

B. Missing else block after if

C. Using parentheses with length property

D. Incorrect use of is operator

Solution

1. Step 1: Check usage of length on String

   In Kotlin, length is a property, not a function, so it should not have parentheses.

2. Step 2: Confirm other parts are correct

   The is operator is used correctly and else block is optional.

3. Final Answer:

   Using parentheses with length property -> Option C

4. Quick Check:

   Property access no parentheses = D [OK]

Hint: Remember: Kotlin properties have no parentheses [OK]

Common Mistakes:

• Adding parentheses to properties like length
• Thinking else block is mandatory
• Misusing is operator syntax

5. Given a list of Any elements, how can you create a new list containing only the strings using the is operator in Kotlin?

hard

A. val strings = list.filterIsInstance<String>()

B. val strings = list.filter { it is String }

C. val strings = list.map { if (it is String) it else null }

D. val strings = list.filter { it == String }

Solution

1. Step 1: Understand filtering by type

   To get only strings, Kotlin provides filterIsInstance<String>() which filters elements by type safely.

2. Step 2: Evaluate other options

   val strings = list.filter { it is String } filters by is String but keeps type as Any, not cast to String. val strings = list.map { if (it is String) it else null } maps to nullable list, not filtering. val strings = list.filter { it == String } compares with type itself, which is invalid.

3. Final Answer:

   val strings = list.filterIsInstance<String>() -> Option A

4. Quick Check:

   Use filterIsInstance for type filtering [OK]

Hint: Use filterIsInstance to get typed list quickly [OK]

Common Mistakes:

• Using filter without casting
• Mapping to nullable instead of filtering
• Comparing element to type object