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Kotlinprogramming~10 mins

Why ranges simplify iteration in Kotlin - Test Your Understanding

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to print numbers from 1 to 5 using a range.

Kotlin
for (i in 1..[1]) {
    println(i)
}
Drag options to blanks, or click blank then click option'
A0
B10
C5
D6
Attempts:
3 left
💡 Hint
Common Mistakes
Using a number outside the desired range like 6 or 10.
Using 0 which is less than the start of the range.
2fill in blank
medium

Complete the code to print numbers from 5 down to 1 using a range.

Kotlin
for (i in [1] downTo 1) {
    println(i)
}
Drag options to blanks, or click blank then click option'
A1
B0
C10
D5
Attempts:
3 left
💡 Hint
Common Mistakes
Using 1 before downTo which would create an empty range.
Using 0 or 10 which are outside the desired range.
3fill in blank
hard

Fix the error in the code to print even numbers from 2 to 10 using a range with step.

Kotlin
for (i in 2..10 step [1]) {
    println(i)
}
Drag options to blanks, or click blank then click option'
A5
B2
C1
D3
Attempts:
3 left
💡 Hint
Common Mistakes
Using step 1 which prints all numbers, not just even ones.
Using step 3 or 5 which skips some even numbers.
4fill in blank
hard

Fill both blanks to create a range that prints numbers from 1 to 10 but only odd numbers.

Kotlin
for (i in [1]..[2] step 2) {
    println(i)
}
Drag options to blanks, or click blank then click option'
A1
B2
C10
D9
Attempts:
3 left
💡 Hint
Common Mistakes
Starting at 2 which would print even numbers.
Ending at 10 which includes an even number and would print 10 if stepped by 2.
5fill in blank
hard

Fill all three blanks to create a range that prints numbers from 10 down to 1, stepping by 3.

Kotlin
for (i in [1] downTo [2] step [3]) {
    println(i)
}
Drag options to blanks, or click blank then click option'
A10
B1
C3
D2
Attempts:
3 left
💡 Hint
Common Mistakes
Using step 2 or 1 which changes the numbers printed.
Swapping start and end values.

Practice

(1/5)
1. What is the main advantage of using ranges in Kotlin for iteration?
easy
A. They let you loop over numbers easily and clearly.
B. They require more code to write loops.
C. They only work with strings, not numbers.
D. They make loops run slower.

Solution

  1. Step 1: Understand what ranges do

    Ranges represent a sequence of numbers, making it easy to loop through them.
  2. Step 2: Compare with other methods

    Using ranges is simpler and clearer than manually managing loop counters.
  3. Final Answer:

    They let you loop over numbers easily and clearly. -> Option A
  4. Quick Check:

    Ranges simplify iteration = They let you loop easily [OK]
Hint: Ranges make number loops simple and clear [OK]
Common Mistakes:
  • Thinking ranges are slower
  • Believing ranges only work with strings
  • Assuming ranges need more code
2. Which of the following is the correct syntax to loop from 1 to 5 using a range in Kotlin?
easy
A. for (i in 1 to 5) { println(i) }
B. for (i in range(1,5)) { println(i) }
C. for (i in 1...5) { println(i) }
D. for (i in 1..5) { println(i) }

Solution

  1. Step 1: Identify Kotlin range syntax

    Kotlin uses two dots '..' to create ranges, like 1..5.
  2. Step 2: Check each option

    for (i in 1..5) { println(i) } uses '1..5' which is correct; others use invalid syntax.
  3. Final Answer:

    for (i in 1..5) { println(i) } -> Option D
  4. Quick Check:

    Correct range syntax = 1..5 [OK]
Hint: Remember Kotlin uses '..' for ranges [OK]
Common Mistakes:
  • Using 'to' instead of '..' for ranges
  • Using three dots '...' which is invalid
  • Trying Python-like range() function
3. What will be the output of this Kotlin code?
for (i in 3..6) { print(i) }
medium
A. 3456
B. 3 4 5 6
C. 3..6
D. Error

Solution

  1. Step 1: Understand the loop range

    The loop runs from 3 to 6 inclusive, so i takes values 3,4,5,6.
  2. Step 2: Analyze the print statement

    print(i) prints numbers without spaces or newlines, so output is '3456'.
  3. Final Answer:

    3456 -> Option A
  4. Quick Check:

    Loop prints numbers directly = 3456 [OK]
Hint: print() outputs without spaces or newlines [OK]
Common Mistakes:
  • Assuming spaces between numbers
  • Thinking '..' prints as string
  • Expecting an error from range
4. Identify the error in this Kotlin code snippet:
for (i in 5..1) { println(i) }
medium
A. Runtime error: cannot loop backwards with '..'.
B. Syntax error: ranges must count up only.
C. The range 5..1 is empty, so loop does not run.
D. No error; it prints 5 to 1.

Solution

  1. Step 1: Check the range direction

    Range 5..1 tries to count up from 5 to 1, which is empty.
  2. Step 2: Understand loop behavior

    Since the range is empty, the loop body never executes, no error occurs.
  3. Final Answer:

    The range 5..1 is empty, so loop does not run. -> Option C
  4. Quick Check:

    Counting up range empty if start > end [OK]
Hint: Ranges count up only unless reversed explicitly [OK]
Common Mistakes:
  • Expecting syntax or runtime error
  • Thinking loop prints numbers backwards
  • Using '..' for descending ranges without 'downTo'
5. How can you use ranges to print numbers from 10 down to 5 in Kotlin?
hard
A. for (i in 10..5) { println(i) }
B. for (i in 10 downTo 5) { println(i) }
C. for (i in 5..10 step -1) { println(i) }
D. for (i in 10..5 step 1) { println(i) }

Solution

  1. Step 1: Understand descending ranges

    To count down, Kotlin uses 'downTo' instead of '..'.
  2. Step 2: Check each option

    for (i in 10 downTo 5) { println(i) } uses '10 downTo 5' which correctly counts down from 10 to 5.
  3. Final Answer:

    for (i in 10 downTo 5) { println(i) } -> Option B
  4. Quick Check:

    Use 'downTo' for descending ranges [OK]
Hint: Use 'downTo' for counting down in ranges [OK]
Common Mistakes:
  • Using '..' for descending ranges
  • Trying negative step in ascending range
  • Confusing step direction
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10 questions~5 mineasy hard