For loop with ranges in Kotlin - Step-by-Step Execution

Concept Flow - For loop with ranges

Start

↓

Set range start..end

↓

For each value in range

↓

Execute loop body

↓

Next value in range?

No→Exit loop

↩Back to Execute loop body

The loop starts by defining a range, then runs the loop body for each value in that range, moving to the next value until the end.

Execution Sample

Kotlin

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

Prints numbers 1, 2, and 3 using a for loop over a range.

Execution Table

Iteration	i	Condition (i in 1..3)	Action	Output
1	1	true	Print i	1
2	2	true	Print i	2
3	3	true	Print i	3
4	4	false	Exit loop

💡 Next value would be 4, which is outside the range 1..3, so the loop stops

Variable Tracker

Variable	Start	After 1	After 2	After 3	Final
i	-	1	2	3	3

Key Moments - 2 Insights

Why does the loop stop before i becomes 4?

Because the range is from 1 to 3, and 4 is outside this range, so the condition i in 1..3 becomes false before assigning i=4 (see execution_table row 4).

Is the last value in the range included in the loop?

Yes, the range 1..3 includes 3, so the loop runs when i is 3 (see execution_table row 3).

Visual Quiz - 3 Questions

Look at the execution table, what is the value of i during the second iteration?

A 2

B 3

C 1

D 4

💡 Hint

Check the 'i' column in the execution_table at iteration 2.

At which iteration does the loop condition become false?

A 3

B 4

C 1

D 2

💡 Hint

Look at the 'Condition' column in the execution_table where it changes to false.

If the range was changed to 1..4, how many times would the loop run?

A 3 times

B 5 times

C 4 times

D 2 times

💡 Hint

The loop runs once for each number in the range, including the last number.

Concept Snapshot

for (variable in start..end) {
    // code to repeat
}

- Runs loop from start to end inclusive
- 'variable' takes each value in the range
- Loop stops after end value
- Simple way to repeat code fixed times

Full Transcript

This visual shows how a Kotlin for loop with a range works. The loop starts by setting a range from 1 to 3. Then, for each number in this range, it runs the loop body which prints the current number. The variable i takes values 1, 2, and 3 in order. After i=3, the next value 4 is outside the range, so the loop stops. This example helps beginners see how the loop variable changes and when the loop ends.

Practice

1. What does the following Kotlin code do?
for (i in 1..5) { println(i) }

easy

A. Prints numbers from 1 to 4

B. Prints numbers from 1 to 5 inclusive

C. Prints numbers from 0 to 5

D. Prints numbers from 5 down to 1

Solution

1. Step 1: Understand the range operator

   The operator 1..5 creates a range including numbers 1, 2, 3, 4, and 5.

2. Step 2: Loop through the range

   The for loop runs from 1 to 5, printing each number.

3. Final Answer:

   Prints numbers from 1 to 5 inclusive -> Option B

4. Quick Check:

   Range 1..5 includes 5 [OK]

Hint: .. means inclusive range from start to end [OK]

Common Mistakes:

• Thinking .. excludes the last number
• Confusing range with step or downTo
• Assuming it counts down automatically

2. Which of the following is the correct syntax for a for loop counting down from 5 to 1 in Kotlin?

easy

A. for (i in 5 downTo 1) { println(i) }

B. for (i in 1..5 step -1) { println(i) }

C. for (i in 5..1) { println(i) }

D. for (i in 1 downTo 5) { println(i) }

Solution

1. Step 1: Identify correct descending range syntax

   Kotlin uses downTo to count down from a higher number to a lower number.

2. Step 2: Check each option

   for (i in 5 downTo 1) { println(i) } uses 5 downTo 1, which is correct. for (i in 5..1) { println(i) } uses an invalid range (5..1). for (i in 1..5 step -1) { println(i) } tries to use negative step incorrectly. for (i in 1 downTo 5) { println(i) } reverses the range incorrectly.

3. Final Answer:

   for (i in 5 downTo 1) { println(i) } -> Option A

4. Quick Check:

   Use downTo for descending loops [OK]

Hint: Use downTo to count down in Kotlin loops [OK]

Common Mistakes:

• Using .. for descending ranges
• Trying negative step with .. (not allowed)
• Swapping start and end in downTo

3. What is the output of this Kotlin code?

var sum = 0
for (i in 1..5 step 2) {
  sum += i
}
println(sum)

medium

A. 6

B. 15

C. 9

D. 10

Solution

1. Step 1: Understand the loop range and step

   The loop runs from 1 to 5, stepping by 2: values are 1, 3, 5.

2. Step 2: Calculate the sum of these values

   sum = 0 + 1 + 3 + 5 = 9

3. Final Answer:

   9 -> Option C

4. Quick Check:

   1 + 3 + 5 = 9 [OK]

Hint: Add numbers in steps to find sum quickly [OK]

Common Mistakes:

• Adding all numbers 1 to 5 instead of stepping
• Using step as a multiplier
• Confusing step with downTo

4. Find the error in this Kotlin code snippet:

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

medium

A. println cannot be used inside for loop

B. step cannot be used with ranges

C. Variable i is not declared

D. Range 10..1 is invalid for counting down

Solution

1. Step 1: Analyze the range 10..1

   The range 10..1 tries to count up from 10 to 1, which is invalid because 10 is greater than 1.

2. Step 2: Understand correct descending syntax

   To count down, use downTo instead of .., e.g., 10 downTo 1 step 2.

3. Final Answer:

   Range 10..1 is invalid for counting down -> Option D

4. Quick Check:

   Use downTo for descending ranges [OK]

Hint: .. always counts up; use downTo to count down [OK]

Common Mistakes:

• Using .. for descending ranges
• Thinking step fixes descending range
• Ignoring Kotlin's range direction rules

5. You want to print all even numbers from 2 to 20 inclusive using a for loop in Kotlin. Which code snippet correctly does this?

hard

A. for (i in 2..20 step 2) { println(i) }

B. for (i in 2..19) { if (i % 2 == 0) println(i) }

C. for (i in 2 downTo 20 step 2) { println(i) }

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

Solution

1. Step 1: Identify the range and step for even numbers

   Even numbers from 2 to 20 can be generated by starting at 2 and stepping by 2 up to 20.

2. Step 2: Check each option

   for (i in 2..20 step 2) { println(i) } uses 2..20 step 2, which prints 2,4,6,...,20 correctly. for (i in 2..19) { if (i % 2 == 0) println(i) } ends at 19 so prints evens up to 18, missing 20. for (i in 2 downTo 20 step 2) { println(i) } counts down incorrectly. for (i in 1..10 step 2) { println(i * 2) } prints 2,6,10,..., not all even numbers.

3. Final Answer:

   for (i in 2..20 step 2) { println(i) } -> Option A

4. Quick Check:

   Use step 2 from 2 to 20 for evens [OK]

Hint: Use step 2 starting at 2 to print even numbers [OK]

Common Mistakes:

• Using downTo instead of .. for ascending
• Forgetting to step by 2
• Multiplying inside loop incorrectly