When with ranges and types in Kotlin - Mini Project: Build & Apply

Using When with Ranges and Types in Kotlin

📖 Scenario: You are building a simple program that categorizes input values based on their type and numeric range. This is useful in many apps where you want to respond differently depending on what kind of data you get and its value.

🎯 Goal: Create a Kotlin program that uses a when expression to check both the type and the numeric range of a variable, then prints a message describing the input.

📋 What You'll Learn

Create a variable called input with a specific value.

Create a variable called ageRange representing ages from 13 to 19 inclusive.

Use a when expression to check the type of input and if it is an Int, check if it falls within ageRange.

Print the correct message based on the checks.

💡 Why This Matters

🌍 Real World

Checking input types and ranges is common in apps that need to validate user data, like age verification or input validation.

💼 Career

Understanding how to use <code>when</code> with types and ranges helps you write clear and concise Kotlin code for decision making, a key skill for Android and backend Kotlin developers.

1

Create the input variable

Create a variable called input and set it to the integer value 16.

Kotlin

# Create a variable called input and set it to 16
# Your code here

Hint

Use val input: Any = 16 to allow different types later.

2

Create the ageRange variable

Create a variable called ageRange that represents the range of integers from 13 to 19 inclusive.

Kotlin

val input: Any = 16
# Create a variable called ageRange for numbers 13 to 19 inclusive
# Your code here

Hint

Use the range operator .. to create the range.

3

Write the when expression to check type and range

Write a when expression using the variable input that checks:
- If input is an Int and is in ageRange, print "Teenager".
- If input is an Int but not in ageRange, print "Not a teenager".
- If input is a String, print "Input is a string".
- For any other type, print "Unknown type".

Kotlin

val input: Any = 16
val ageRange = 13..19

// Write a when expression to check input type and range
// Your code here

Hint

Use is to check type and in to check if number is in range.

4

Print the result

Run the program to print the correct message based on the input value and the when expression.

Kotlin

val input: Any = 16
val ageRange = 13..19

when (input) {
    is Int -> if (input in ageRange) {
        println("Teenager")
    } else {
        println("Not a teenager")
    }
    is String -> println("Input is a string")
    else -> println("Unknown type")
}

// Run the program to see the output
# Your code here

Hint

Running the program should print "Teenager" because 16 is in the range 13 to 19.

Practice

1. What does the when expression in Kotlin do when used with ranges like in 1..5?

easy

A. It throws an error if the value is outside the range.

B. It creates a new range from 1 to 5.

C. It assigns the value to the range variable.

D. It checks if the value falls within the specified range.

Solution

1. Step 1: Understand the when with range syntax

   The syntax in 1..5 inside when checks if the value is between 1 and 5 inclusive.

2. Step 2: Interpret the check

   If the value is inside the range, the corresponding branch executes; otherwise, it moves to the next case.

3. Final Answer:

   It checks if the value falls within the specified range. -> Option D

4. Quick Check:

   when with in checks range membership [OK]

Hint: Remember: in means inside the range [OK]

Common Mistakes:

• Thinking in creates a range instead of checking membership
• Confusing in with assignment
• Assuming it throws errors for out-of-range values

2. Which of the following is the correct syntax to check if a variable x is NOT in the range 10 to 20 using when in Kotlin?

easy

A. when(x) { !in 10..20 -> "Outside" }

B. when(x) { not in 10..20 -> "Outside" }

C. when(x) { x !in 10..20 -> "Outside" }

D. when(x) { outside 10..20 -> "Outside" }

Solution

1. Step 1: Recall the correct negation syntax in when

   In Kotlin, !in is used to check if a value is NOT in a range.

2. Step 2: Match the syntax with options

   when(x) { !in 10..20 -> "Outside" } uses !in 10..20 correctly inside when. Other options use invalid keywords or syntax.

3. Final Answer:

   when(x) { !in 10..20 -> "Outside" } -> Option A

4. Quick Check:

   Use !in to check exclusion from range [OK]

Hint: Use !in to say NOT in range [OK]

Common Mistakes:

• Using 'not in' instead of '!in'
• Writing variable name inside when case
• Using undefined keywords like 'outside'

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

val x: Any = 42
val result = when (x) {
  is Int -> "Integer"
  in 1..10 -> "Small number"
  else -> "Unknown"
}
println(result)

medium

A. Small number

B. Unknown

C. Integer

D. Compilation error

Solution

1. Step 1: Check the type of x

   Variable x is of type Any but holds the value 42, which is an Int.

2. Step 2: Evaluate when branches in order

   The first branch checks is Int, which is true, so it returns "Integer". The second branch is not checked because the first matched.

3. Final Answer:

   Integer -> Option C

4. Quick Check:

   when matches first true branch [OK]

Hint: Type checks with is come before range checks [OK]

Common Mistakes:

• Assuming range check runs before type check
• Confusing value 42 as outside 1..10
• Thinking when checks all branches

4. Identify the error in this Kotlin when expression:

val y: Any = "Hello"
val output = when (y) {
  in 1..5 -> "Number"
  is String -> "Text"
  else -> "Other"
}
println(output)

medium

A. Cannot use in with Any type without casting

B. The is String check should come before the range check

C. Missing break statements in when

D. No error, code runs fine

Solution

1. Step 1: Analyze the in 1..5 check on Any

   The variable y is of type Any. The in operator with ranges requires a comparable type like Int. Using in 1..5 on Any causes a type mismatch error.

2. Step 2: Check order of branches and syntax

   The is String check is valid but unreachable if the first branch errors. Also, when in Kotlin does not require break.

3. Final Answer:

   Cannot use in with Any type without casting -> Option A

4. Quick Check:

   Range checks need compatible types [OK]

Hint: Check variable type before using in with ranges [OK]

Common Mistakes:

• Ignoring type mismatch for in operator
• Thinking when needs break
• Assuming order of branches causes error

5. You want to write a Kotlin when expression that returns:
- "Small Int" if the value is an Int between 1 and 10 inclusive,
- "Large Int" if the value is an Int greater than 10,
- "String" if the value is a String,
- "Other" for anything else.

Which of the following when expressions correctly implements this?

hard

A. when (v) {
in 1..10 -> "Small Int"
is Int -> "Large Int"
is String -> "String"
else -> "Other"
}

B. when (v) {
is Int -> if (v in 1..10) "Small Int" else "Large Int"
is String -> "String"
else -> "Other"
}

C. when (v) {
is Int in 1..10 -> "Small Int"
is Int -> "Large Int"
is String -> "String"
else -> "Other"
}

D. when (v) {
is Int && v in 1..10 -> "Small Int"
is Int -> "Large Int"
is String -> "String"
else -> "Other"
}

Solution

1. Step 1: Understand the conditions needed

   We need to check if v is an Int and then check its value range for "Small Int" or "Large Int". Also check if v is a String.

2. Step 2: Analyze each option's correctness

   when (v) { in 1..10 -> "Small Int" is Int -> "Large Int" is String -> "String" else -> "Other" } incorrectly uses in 1..10 without checking type first, causing errors if v is not Int. when (v) { is Int in 1..10 -> "Small Int" is Int -> "Large Int" is String -> "String" else -> "Other" } uses invalid syntax is Int in 1..10. when (v) { is Int && v in 1..10 -> "Small Int" is Int -> "Large Int" is String -> "String" else -> "Other" } uses invalid syntax is Int && v in 1..10 inside when. when (v) { is Int -> if (v in 1..10) "Small Int" else "Large Int" is String -> "String" else -> "Other" } correctly checks type first, then uses an if inside the branch to distinguish ranges.

3. Final Answer:

   Option B correctly implements the logic. -> Option B

4. Quick Check:

   Use nested if inside when for complex conditions [OK]

Hint: Check type first, then range inside branch with if [OK]

Common Mistakes:

• Using invalid syntax combining is and in
• Not checking type before range
• Trying to use logical operators directly in when cases