For loop with step and downTo in Kotlin - Mini Project: Build & Apply

For loop with step and downTo in Kotlin

📖 Scenario: You are organizing a countdown event and want to display numbers counting down with a specific step.

🎯 Goal: Build a Kotlin program that counts down from 10 to 1, stepping by 2, and prints each number.

📋 What You'll Learn

Create a variable called start with the value 10

Create a variable called end with the value 1

Use a for loop with downTo and step to count down from start to end by 2

Print each number inside the loop

💡 Why This Matters

🌍 Real World

Counting down with steps is useful in timers, animations, and game loops where you want to skip certain numbers.

💼 Career

Understanding loops with steps and counting down is important for writing efficient code in many programming jobs, including app development and automation.

1

Create start and end variables

Create a variable called start and set it to 10. Create another variable called end and set it to 1.

Kotlin

# Create variables start and end
// Your code here

Hint

Use val to create variables and assign the numbers 10 and 1.

2

Write a for loop with downTo and step

Write a for loop using i as the loop variable that counts down from start to end using downTo and steps by 2 using step.

Kotlin

val start = 10
val end = 1

// Write a for loop counting down from start to end by 2
// Your code here

Hint

Use for (i in start downTo end step 2) to count down by 2.

3

Print each number inside the loop

Inside the for loop, write println(i) to print each number.

Kotlin

val start = 10
val end = 1

for (i in start downTo end step 2) {
    // Print each number
    // Your code here
}

Hint

Use println(i) inside the loop to show each number.

4

Run the program to see the countdown

Run the program to print the countdown numbers from 10 to 1 stepping by 2.

Kotlin

val start = 10
val end = 1

for (i in start downTo end step 2) {
    println(i)
}

// Run the program to see output

Hint

The program should print numbers 10, 8, 6, 4, and 2 each on a new line.

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