until loop in Bash Scripting - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create an until loop that stops when the variable count reaches 5.

Bash Scripting

count=1
until [ $count [1] 5 ]
do
  echo $count
  ((count++))
done

Drag options to blanks, or click blank then click option'

A-lt

B-gt

C-eq

D-ne

Attempts:

3 left

💡 Hint

Common Mistakes

Using -lt instead of -gt causes the loop to never run.

Using -eq stops the loop early.

The until loop runs until the condition is true. Here, it runs while count is less than 5.

2fill in blank

medium

Complete the code to increment the variable inside the until loop.

Bash Scripting

num=0
until [ $num -eq 3 ]
do
  echo $num
  [1]
done

Drag options to blanks, or click blank then click option'

A((num++))

Bnum += 1

Cnum=num+1

Dnum++

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'num=num+1' treats num as a string, not a number.

Using 'num++' alone is not valid in bash.

In bash, arithmetic expressions like ((num++)) correctly increment the variable.

3fill in blank

hard

Fix the error in the until loop condition to stop when var equals 10.

Bash Scripting

var=0
until [ $var [1] 10 ]
do
  echo $var
  ((var+=2))
done

Drag options to blanks, or click blank then click option'

A-lt

B-le

C-gt

D-ne

Attempts:

3 left

💡 Hint

Common Mistakes

Using -lt stops the loop too early.

Using -gt causes the loop to never run.

The until loop runs until the condition is true. Using -ne means it runs while var is not equal to 10.

4fill in blank

hard

Fill both blanks to create an until loop that prints numbers from 1 to 4.

Bash Scripting

i=1
until [ $i [1] 5 ]
do
  echo $i
  [2]
done

Drag options to blanks, or click blank then click option'

A-eq

B((i++))

C-lt

Di+=1

Attempts:

3 left

💡 Hint

Common Mistakes

Using -lt stops the loop immediately.

Using 'i+=1' alone is not valid syntax in bash.

The loop runs until i equals 5, so it runs while i is less than 5. Incrementing i with ((i++)) updates the variable correctly.

5fill in blank

hard

Fill all three blanks to create an until loop that prints even numbers from 2 to 8.

Bash Scripting

num=2
until [ $num [1] 10 ]
do
  echo $num
  [2]
done

Drag options to blanks, or click blank then click option'

A-ge

B((num+=2))

C-lt

Dnum++

Attempts:

3 left

💡 Hint

Common Mistakes

Using -ge stops the loop immediately.

Using num++ increments by 1, not 2.

The loop runs while num is less than 10. Incrementing num by 2 each time prints even numbers.

Practice

1. What does an until loop do in bash scripting?

easy

A. It repeats commands until the condition becomes true.

B. It repeats commands while the condition is true.

C. It runs commands only once regardless of condition.

D. It runs commands only if the condition is false.

Solution

1. Step 1: Understand the until loop behavior

   An until loop runs repeatedly until its condition becomes true, meaning it stops when the condition is true.

2. Step 2: Compare options with definition

   It repeats commands until the condition becomes true. correctly states the loop repeats until the condition is true. It repeats commands while the condition is true. describes a while loop. Options C and D are incorrect behaviors.

3. Final Answer:

   It repeats commands until the condition becomes true. -> Option A

4. Quick Check:

   until loop = repeat until true [OK]

Hint: Remember: until stops when condition is true [OK]

Common Mistakes:

• Confusing until with while loop
• Thinking until runs only once
• Believing until runs while condition is true

2. Which of the following is the correct syntax for an until loop in bash?

easy

A. until [ condition ]; do commands; done

B. until ( condition ) { commands }

C. until { condition } do commands done

D. until condition do commands done

Solution

1. Step 1: Recall bash until loop syntax

   The correct syntax uses square brackets for condition, followed by do and done keywords: until [ condition ]; do commands; done

2. Step 2: Check each option

   until [ condition ]; do commands; done matches correct syntax. Options B, C, and D use incorrect brackets or missing semicolons.

3. Final Answer:

   until [ condition ]; do commands; done -> Option A

4. Quick Check:

   Syntax = until [ condition ]; do ... done [OK]

Hint: Use square brackets and do/done keywords [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Omitting do or done keywords
• Missing semicolon before do

3. What will be the output of this script?

count=1
until [ $count -gt 3 ]
do
  echo $count
  ((count++))
done

medium

A. 1 2

B. 1 2 3 4

C. 1 2 3

D. No output, infinite loop

Solution

1. Step 1: Analyze loop condition and increments

   The loop runs until count is greater than 3. Starting at 1, it prints count and increments it each time.

2. Step 2: Trace iterations

   count=1 (prints 1), count=2 (prints 2), count=3 (prints 3), count=4 stops loop because 4 > 3.

3. Final Answer:

   1 2 3 -> Option C

4. Quick Check:

   Prints 1 to 3 before stopping [OK]

Hint: Loop stops when count > 3, so prints 1 to 3 [OK]

Common Mistakes:

• Thinking it prints 4
• Confusing until with while loop condition
• Assuming infinite loop

4. Identify the error in this until loop script:

count=1
until [ $count -lt 5 ]
do
  echo $count
  ((count++))
done

medium

A. Missing semicolon after condition.

B. The loop will never run because condition is initially true.

C. The variable count is not initialized.

D. The condition should use -gt instead of -lt.

Solution

1. Step 1: Understand the until loop condition

   until runs while the condition is false and stops when true. Here, condition is count -lt 5, which is true at start (count=1).

2. Step 2: Analyze loop execution

   Since condition is true initially, loop does not run even once.

3. Final Answer:

   The loop will never run because condition is initially true. -> Option B

4. Quick Check:

   until stops immediately if condition true [OK]

Hint: until runs only while condition is false [OK]

Common Mistakes:

• Confusing -lt with -gt in condition
• Assuming loop runs at least once
• Missing that until stops when condition true

5. You want to write a bash script that waits until a file named ready.txt exists before continuing. Which until loop correctly implements this?

hard

A. until [ ! -f ready.txt ]; do sleep 1; done

B. until [ ! -f ready.txt ]; do echo 'Waiting'; done

C. until [ -d ready.txt ]; do sleep 1; done

D. until [ -f ready.txt ]; do sleep 1; done

Solution

1. Step 1: Understand the goal and condition

   The script should wait until the file exists. The loop runs while the condition is false, so condition should be "file exists" [ -f ready.txt ].

2. Step 2: Check each option's condition

   until [ -f ready.txt ]; do sleep 1; done loops while [ -f ] is false (file missing), sleeping until file exists. B, C (-d for directory), and D use wrong conditions.

3. Final Answer:

   until [ -f ready.txt ]; do sleep 1; done -> Option D

4. Quick Check:

   Loop while file missing, stops when exists [OK]

Hint: Condition [ -f file ] stops when file exists [OK]

Common Mistakes:

• Using -d instead of -f for file check
• Using ! -f negation incorrectly
• Not including sleep to avoid busy wait