Range operator (..) and in operator in Kotlin

Introduction

The range operator (..) helps create a sequence of values easily. The in operator checks if a value is inside that sequence.

When you want to repeat an action a certain number of times.

When you want to check if a number falls between two limits.

When you want to loop through a list of numbers or characters.

When you want to verify if a value exists within a range.

When you want to simplify conditions that check for value boundaries.

Syntax

Kotlin

val range = start..end
if (value in range) {
    // do something
}

The .. operator creates a range from start to end, including both ends.

The in operator checks if a value is inside the range or collection.

Examples

This creates a range from 1 to 5 and prints the list [1, 2, 3, 4, 5].

Kotlin

val numbers = 1..5
println(numbers.toList())

This checks if the letter 'c' is inside the range from 'a' to 'd'. It prints true.

Kotlin

val letterRange = 'a'..'d'
println('c' in letterRange)

This loops from 1 to 3 and prints each number on its own line.

Kotlin

for (i in 1..3) {
    println(i)
}

Sample Program

This program checks if a score is within the passing range using the in operator. Then it prints numbers from 1 to 5 using the range operator in a loop.

Kotlin

fun main() {
    val score = 85
    val passingRange = 60..100

    if (score in passingRange) {
        println("You passed with a score of $score.")
    } else {
        println("You failed with a score of $score.")
    }

    println("Numbers from 1 to 5:")
    for (num in 1..5) {
        print("$num ")
    }
}

Important Notes

The range operator works with numbers and characters.

Ranges include both the start and end values.

You can use !in to check if a value is NOT in a range.

Summary

The .. operator creates a range of values.

The in operator checks if a value is inside that range.

Ranges are useful for loops and condition checks.

Practice

1. What does the Kotlin expression 1..5 represent?

easy

A. An error because ranges need parentheses

B. A list containing numbers 1 and 5 only

C. A function that returns 1 or 5

D. A range of numbers from 1 to 5 inclusive

Solution

1. Step 1: Understand the range operator

   The .. operator creates a range including both start and end values.

2. Step 2: Interpret 1..5

   This means all numbers from 1 up to 5, including 1 and 5.

3. Final Answer:

   A range of numbers from 1 to 5 inclusive -> Option D

4. Quick Check:

   1..5 = numbers 1 to 5 [OK]

Hint: Remember: .. includes both ends in Kotlin ranges [OK]

Common Mistakes:

• Thinking it creates a list instead of a range
• Assuming it excludes the last number
• Confusing it with a function call

2. Which of the following is the correct syntax to check if variable x is between 10 and 20 inclusive in Kotlin?

easy

A. if (x between 10 and 20) { }

B. if (x in 10..20) { }

C. if (10 <= x <= 20) { }

D. if (x inside 10..20) { }

Solution

1. Step 1: Recall Kotlin's in operator syntax

   Kotlin uses in to check if a value is inside a range.

2. Step 2: Match syntax with options

   Only if (x in 10..20) correctly uses in and the range operator ...

3. Final Answer:

   if (x in 10..20) { } -> Option B

4. Quick Check:

   Use in with .. for range checks [OK]

Hint: Use 'in' with '..' for range checks in Kotlin [OK]

Common Mistakes:

• Using 'between' or 'inside' which are not Kotlin keywords
• Trying chained comparisons like 10 <= x <= 20
• Missing the range operator '..'

3. What is the output of this Kotlin code?

val x = 7
if (x in 1..5) println("Inside") else println("Outside")

medium

A. Outside

B. Compilation error

C. Inside

D. No output

Solution

1. Step 1: Understand the range check

   The range 1..5 includes numbers 1, 2, 3, 4, 5.

2. Step 2: Check if x=7 is in the range

   7 is not between 1 and 5, so the condition is false.

3. Final Answer:

   Outside -> Option A

4. Quick Check:

   7 not in 1..5 means print "Outside" [OK]

Hint: Check if value lies within range boundaries [OK]

Common Mistakes:

• Assuming 7 is inside 1..5
• Confusing inclusive ranges with exclusive
• Expecting a syntax error

4. Identify the error in this Kotlin code snippet:

for (i in 1..5)
    if (i in 3..1) println(i)

medium

A. The for loop syntax is incorrect

B. The range 3..1 is invalid because start is greater than end

C. No error, code runs fine

D. The if condition should use 'not in' instead of 'in'

Solution

1. Step 1: Analyze the range 3..1

   In Kotlin, a range with start greater than end (3..1) creates an empty range, which is valid.

2. Step 2: Verify code execution

   The for loop runs for i=1 to 5, but i in 3..1 is always false for any i, so nothing prints. No syntax or runtime errors occur.

3. Final Answer:

   No error, code runs fine -> Option C

4. Quick Check:

   Empty ranges (start > end) are valid in Kotlin [OK]

Hint: Ranges with start > end are empty but valid [OK]

Common Mistakes:

• Thinking reversed ranges like 3..1 are invalid
• Assuming the for loop syntax is wrong
• Expecting a syntax error or crash with empty range

5. You want to print all even numbers between 2 and 10 inclusive using a Kotlin for loop and range. Which code snippet correctly does this?

hard

A. for (i in 2..10 step 2) println(i)

B. for (i in 2..10) if (i % 2 != 0) println(i)

C. for (i in 2 until 10 step 2) println(i)

D. for (i in 2..10 step 3) println(i)

Solution

1. Step 1: Understand the range and step usage

   2..10 step 2 creates numbers 2, 4, 6, 8, 10.

2. Step 2: Check each option

   for (i in 2..10 step 2) println(i) prints 2,4,6,8,10 directly. for (i in 2..10) if (i % 2 != 0) println(i) checks i % 2 != 0 and prints odds 3,5,7,9. for (i in 2 until 10 step 2) println(i) uses until excluding 10 (2,4,6,8). for (i in 2..10 step 3) println(i) steps by 3 (2,5,8).

3. Final Answer:

   for (i in 2..10 step 2) println(i) -> Option A

4. Quick Check:

   Use step 2 with inclusive range for even numbers [OK]

Hint: Use 'step 2' with '..' to get even numbers directly [OK]

Common Mistakes:

• Using wrong modulo condition like % 2 != 0
• Using 'until' which excludes the end value
• Using wrong step size like 3