For loop with step and downTo in Kotlin

Introduction

A for loop with step and downTo helps you count up or down by a specific amount, making it easy to repeat actions in a controlled way.

When you want to count backwards through numbers, like counting down a timer.

When you want to skip numbers while looping, like checking every second item in a list.

When you want to print numbers in reverse order with a fixed gap.

When you need to perform an action a certain number of times but not one by one.

Syntax

Kotlin

for (i in start downTo end step stepValue) {
    // code to repeat
}

Use downTo to count backwards from start to end.

Use step to skip numbers by the stepValue.

Examples

Counts down from 10 to 1, skipping every other number.

Kotlin

for (i in 10 downTo 1 step 2) {
    println(i)
}

Counts down from 5 to 0 by 1 each time.

Kotlin

for (i in 5 downTo 0) {
    println(i)
}

Counts down from 20 to 10, jumping 3 numbers each time.

Kotlin

for (i in 20 downTo 10 step 3) {
    println(i)
}

Sample Program

This program prints numbers from 10 down to 1, skipping every second number.

Kotlin

fun main() {
    println("Counting down from 10 to 1, stepping by 2:")
    for (i in 10 downTo 1 step 2) {
        println(i)
    }
}

Important Notes

If you omit step, the loop counts down by 1 by default.

The downTo keyword always counts backwards, so start should be greater than or equal to end.

Using step with a value less than 1 will cause an error.

Summary

Use downTo to count backwards in a for loop.

Use step to skip numbers while looping.

Together, they let you control how the loop counts down.

Practice

1. What does the downTo keyword do in a Kotlin for loop?

easy

A. It makes the loop count forwards from a lower number to a higher number.

B. It makes the loop count backwards from a higher number to a lower number.

C. It skips numbers in the loop by a fixed step.

D. It stops the loop immediately.

Solution

1. Step 1: Understand the purpose of downTo

   The downTo keyword in Kotlin is used to create a range that counts backwards, from a higher number down to a lower number.

2. Step 2: Compare with other options

   It makes the loop count forwards from a lower number to a higher number. describes counting forwards, It skips numbers in the loop by a fixed step. skipping numbers by step, and It stops the loop immediately. stopping the loop immediately, none of which downTo does.

3. Final Answer:

   It makes the loop count backwards from a higher number to a lower number. -> Option B

4. Quick Check:

   downTo = count backwards [OK]

Hint: Remember: downTo means counting downwards [OK]

Common Mistakes:

• Confusing downTo with counting forwards
• Thinking downTo skips numbers
• Assuming downTo stops the loop

2. Which of the following is the correct syntax to loop from 10 down to 1, stepping by 2 in Kotlin?

easy

A. for (i in 10..1 step 2) { println(i) }

B. for (i in 1 downTo 10 step 2) { println(i) }

C. for (i in 10 downTo 1 step 2) { println(i) }

D. for (i in 10 downTo 1 by 2) { println(i) }

Solution

1. Step 1: Identify correct downTo and step syntax

   The correct syntax to count down with steps is for (i in start downTo end step stepValue). for (i in 10 downTo 1 step 2) { println(i) } matches this.

2. Step 2: Check other options for errors

   for (i in 1 downTo 10 step 2) { println(i) } has start < end, resulting in empty range. for (i in 10..1 step 2) { println(i) } uses .. which doesn't support down with step correctly. for (i in 10 downTo 1 by 2) { println(i) } uses by instead of step, which is incorrect.

3. Final Answer:

   for (i in 10 downTo 1 step 2) { println(i) } -> Option C

4. Quick Check:

   Correct syntax = for (i in 10 downTo 1 step 2) { println(i) } [OK]

Hint: Use downTo with step, not by [OK]

Common Mistakes:

• Using by instead of step
• Counting down with .. range
• Reversing start and end values

3. What is the output of the following Kotlin code?

for (i in 8 downTo 2 step 3) {
    print("$i ")
}

medium

A. 8 5 2

B. 8 6 4 2

C. 8 5 3

D. 8 5 2 1

Solution

1. Step 1: Determine the loop values

   The loop starts at 8, counts down to 2, stepping by 3. So values are 8, then 8-3=5, then 5-3=2.

2. Step 2: Confirm the loop stops at 2

   Since 2 is the end value and included, the loop prints 8, 5, and 2.

3. Final Answer:

   8 5 2 -> Option A

4. Quick Check:

   8, 5, 2 with step 3 downTo 2 = 8 5 2 [OK]

Hint: Subtract step each time, stop at end value [OK]

Common Mistakes:

• Including numbers less than end value
• Using wrong step size
• Confusing downTo with ..

4. Identify the error in this Kotlin loop:

for (i in 5 downTo 1 step 0) {
    println(i)
}

medium

A. The downTo keyword is misspelled.

B. There is no error; the code runs fine.

C. The loop should use .. instead of downTo.

D. Step value cannot be zero; it causes a runtime error.

Solution

1. Step 1: Analyze the step value

   The step value is 0, which is invalid and causes a runtime error (IllegalArgumentException).

2. Step 2: Check other options

   downTo is spelled correctly, and using .. would count upwards, not downwards. So options A, C, and D are incorrect.

3. Final Answer:

   Step value cannot be zero; it causes a runtime error. -> Option D

4. Quick Check:

   Step must be positive and non-zero [OK]

Hint: Step cannot be zero; it must be positive [OK]

Common Mistakes:

• Using zero or negative step incorrectly
• Confusing downTo with ..
• Ignoring infinite loop risk

5. You want to print all odd numbers from 15 down to 1 using a Kotlin for loop with downTo and step, without using an if statement. Which code snippet correctly does this?

hard

A. for (i in 15 downTo 1 step 2) { println(i) }

B. for (i in 15 downTo 1 step 1) { if (i % 2 == 1) println(i) }

C. for (i in 15 downTo 1 step 3) { println(i) }

D. for (i in 14 downTo 1 step 2) { println(i) }

Solution

1. Step 1: Understand the goal

   We want to print odd numbers from 15 down to 1. Starting at 15 (which is odd) and stepping by 2 will print all odd numbers downwards.

2. Step 2: Evaluate options

   for (i in 15 downTo 1 step 2) { println(i) } starts at 15 and steps by 2 down to 1, printing 15, 13, 11, ..., 1. for (i in 15 downTo 1 step 1) { if (i % 2 == 1) println(i) } uses step 1 with an if filter (disallowed). for (i in 15 downTo 1 step 3) { println(i) } steps by 3, skipping some odds. for (i in 14 downTo 1 step 2) { println(i) } starts at 14 (even), printing evens.

3. Final Answer:

   for (i in 15 downTo 1 step 2) { println(i) } -> Option A

4. Quick Check:

   Start odd, step 2 downTo 1 = odd numbers [OK]

Hint: Start at odd number, step 2 downTo 1 for odd numbers [OK]

Common Mistakes:

• Starting at even number when printing odds
• Using wrong step size
• Filtering inside loop instead of stepping