while loop in Bash Scripting - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

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

Bash Scripting

#!/bin/bash
count=1
while [ $count [1] 6 ]
do
  echo $count
  ((count++))
done

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

A-eq

B-gt

C-lt

D-ne

Attempts:

3 left

💡 Hint

Common Mistakes

Using -gt (greater than) will cause the loop not to run as expected.

Using -eq (equal) will only run if count equals 5, which is not the goal.

The operator -lt means "less than" in bash. The loop runs while count is less than 6.

2fill in blank

medium

Complete the code to stop the loop when the user enters 'exit'.

Bash Scripting

#!/bin/bash
input=""
while [ "$input" != "exit" ]
do
  read -p "Type something (or 'exit' to quit): " [1]
done

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

Aanswer

Btext

Cline

Dinput

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable name than the one in the condition causes the loop to never end.

Forgetting to quote the variable in the condition can cause errors.

The variable input stores what the user types. The loop checks if input is not 'exit'.

3fill in blank

hard

Fix the error in the loop condition to correctly check if count is less than or equal to 10.

Bash Scripting

#!/bin/bash
count=1
while [ $count [1] 10 ]
do
  echo $count
  ((count++))
done

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

A-le

B-eq

C-lt

D-ge

Attempts:

3 left

💡 Hint

Common Mistakes

Using -lt will stop the loop before count reaches 10.

Using -ge or -eq will cause the loop to behave incorrectly.

The operator -le means "less than or equal to" in bash, which matches the requirement.

4fill in blank

hard

Fill both blanks to create a loop that prints even numbers from 2 to 10.

Bash Scripting

#!/bin/bash
num=2
while [ $num [1] 10 ]
do
  echo $num
  num=$((num [2] 2))
done

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

A-le

B+

C-

D-lt

Attempts:

3 left

💡 Hint

Common Mistakes

Using -lt will exclude 10 from the output.

Subtracting 2 will cause an infinite loop.

The loop runs while num is less than or equal to 10 (-le). We add 2 each time (+) to get even numbers.

5fill in blank

hard

Fill all three blanks to create a loop that counts down from 5 to 1 and prints each number.

Bash Scripting

#!/bin/bash
count=5
while [ $count [1] 0 ]
do
  echo [2]
  count=[3]
done

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

A-gt

B$count

C$((count - 1))

D-lt

Attempts:

3 left

💡 Hint

Common Mistakes

Using -lt will cause the loop to never run.

Not decreasing count will cause an infinite loop.

The loop runs while count is greater than 0 (-gt). We print $count and then subtract 1 each time ($((count - 1))).

Practice

1. What does a while loop do in a bash script?

easy

A. Stops the script immediately

B. Runs commands only once

C. Executes commands only if a condition is false

D. Repeats commands as long as a condition is true

Solution

1. Step 1: Understand the purpose of a while loop

   A while loop runs repeatedly while its condition remains true.

2. Step 2: Compare options with the definition

   Only Repeats commands as long as a condition is true correctly describes this behavior.

3. Final Answer:

   Repeats commands as long as a condition is true -> Option D

4. Quick Check:

   while loop = repeat while true [OK]

Hint: While loops run repeatedly while condition is true [OK]

Common Mistakes:

• Thinking while loops run only once
• Confusing while with if statements
• Assuming it runs when condition is false

2. Which of the following is the correct syntax to start a while loop in bash?

easy

A. while [ condition ]; do

B. while ( condition ) {

C. while condition then

D. while { condition } do

Solution

1. Step 1: Recall bash while loop syntax

   The correct syntax uses square brackets for condition and 'do' keyword.

2. Step 2: Match options with correct syntax

   while [ condition ]; do matches the correct bash syntax: while [ condition ]; do.

3. Final Answer:

   while [ condition ]; do -> Option A

4. Quick Check:

   Correct bash while syntax = while [ condition ]; do [OK]

Hint: Use square brackets and 'do' to start while loop [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Missing 'do' keyword
• Using curly braces incorrectly

3. What will be the output of this script?

count=1
while [ $count -le 3 ]; do
  echo $count
  count=$((count + 1))
done

medium

A. 1 2 3 4

B. 1 2 3

C. 3 2 1

D. No output

Solution

1. Step 1: Analyze the loop condition and increments

   The loop starts with count=1 and runs while count is less or equal to 3, incrementing count by 1 each time.

2. Step 2: Trace the output for each iteration

   It prints 1, then 2, then 3, then stops because count becomes 4.

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

   Loop prints 1 to 3 [OK]

Hint: Count increments until condition false, prints each value [OK]

Common Mistakes:

• Forgetting to increment count causing infinite loop
• Expecting output beyond 3
• Confusing loop condition operators

4. Identify the error in this script:

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

medium

A. Wrong comparison operator '-lt'

B. Incorrect variable increment syntax

C. Missing 'do' keyword after while condition

D. Missing closing bracket in condition

Solution

1. Step 1: Check while loop syntax

   The while loop must have 'do' after the condition to start the loop body.

2. Step 2: Identify missing 'do'

   The script misses 'do' after the condition line, causing syntax error.

3. Final Answer:

   Missing 'do' keyword after while condition -> Option C

4. Quick Check:

   while loop needs 'do' [OK]

Hint: Always put 'do' after while condition [OK]

Common Mistakes:

• Omitting 'do' keyword
• Misplacing brackets
• Incorrect increment syntax

5. You want to write a bash script that prints all even numbers from 2 to 10 using a while loop. Which script correctly does this?

hard

A. num=1
while [ $num -le 10 ]; do
if [ $((num % 2)) -eq 0 ]; then echo $num; fi
num=$((num + 1))
done

B. num=2
while [ $num -le 10 ]; do
echo $num
num=$((num + 1))
done

C. num=2
while [ $num -lt 10 ]; do
echo $num
num=$((num + 2))
done

D. num=0
while [ $num -le 10 ]; do
echo $num
num=$((num + 2))
done

Solution

1. Step 1: Understand the goal

   We want to print even numbers from 2 to 10 inclusive.

2. Step 2: Check each option's logic

   num=2 while [ $num -le 10 ]; do echo $num num=$((num + 1)) done starts at 2 and increments by 1 up to 10, printing 2 3 4 5 6 7 8 9 10 - incorrect.
   num=1 while [ $num -le 10 ]; do if [ $((num % 2)) -eq 0 ]; then echo $num; fi num=$((num + 1)) done starts at 1, increments by 1, and prints only if even - correct.
   num=2 while [ $num -lt 10 ]; do echo $num num=$((num + 2)) done stops before 10, missing 10.
   num=0 while [ $num -le 10 ]; do echo $num num=$((num + 2)) done starts at 0, which is even but outside requested range (2 to 10).

3. Step 3: Choose the best option

   Only the script using modulo to explicitly check evenness prints exactly 2,4,6,8,10.

4. Final Answer:

   num=1 while [ $num -le 10 ]; do if [ $((num % 2)) -eq 0 ]; then echo $num; fi num=$((num + 1)) done -> Option A

5. Quick Check:

   Check even with modulo inside loop [OK]

Hint: Use modulo inside loop to print even numbers [OK]

Common Mistakes:

• Stopping loop before 10
• Starting from 0 when 2 is required
• Not checking evenness explicitly