while loop in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

Output of a simple while loop counting down

What is the output of this Bash script?

Bash Scripting

count=3
while [ $count -gt 0 ]
do
  echo $count
  count=$((count - 1))
done

A

3
2
1

B

3
2
1
0

C

1
2
3

D

0
1
2
3

💡 Hint

Remember the loop runs while count is greater than zero and decreases count each time.

The loop starts with count=3 and prints it, then decreases count by 1 until count is no longer greater than 0. So it prints 3, 2, 1.

💻 Command Output

intermediate

While loop with a break condition

What will this Bash script output?

Bash Scripting

i=1
while true
 do
  echo $i
  if [ $i -eq 3 ]; then
    break
  fi
  i=$((i + 1))
done

A

1
2

B

1
2
3
4

C

1
2
3

D

1
2
3
4
5

💡 Hint

The loop breaks when i equals 3.

The loop prints i starting from 1, increments i, and breaks when i reaches 3, so it prints 1, 2, 3.

📝 Syntax

advanced

Identify the syntax error in this while loop

Which option shows the correct syntax to fix the error in this Bash while loop?

Bash Scripting

count=5
while [ $count -gt 0 ]
  echo $count
  count=$((count - 1))
done

A

Add 'do' after the while condition:
while [ $count -gt 0 ]; do

B Replace 'done' with 'end'

C Remove the square brackets from the condition

D Add a semicolon after 'done'

💡 Hint

A while loop in Bash requires a 'do' keyword after the condition.

The 'do' keyword is mandatory after the while condition to start the loop body. Without it, the script has a syntax error.

🔧 Debug

advanced

Why does this while loop run infinitely?

This script runs forever. What is the reason?

Bash Scripting

num=1
while [ $num -le 3 ]
do
  echo $num
done

A The echo command causes the loop to restart

B Variable 'num' is never updated inside the loop

C Missing 'do' keyword after while condition

D The condition uses '-le' which is invalid

💡 Hint

Check if the loop changes the variable controlling the condition.

The variable 'num' is never incremented, so the condition is always true, causing an infinite loop.

🚀 Application

expert

Create a while loop to read lines from a file

Which script correctly reads each line from 'input.txt' and prints it?

A

while read line; do
  echo $line
  done > input.txt

B

while [ read line ]; do
  echo $line
done < input.txt

C

while read line
  echo $line
done < input.txt

D

while read line; do
  echo "$line"
done < input.txt

💡 Hint

The 'read' command reads a line, and the loop must redirect input from the file.

Option D correctly uses 'while read line; do ... done < input.txt' to read and print each line. Others have syntax errors or wrong redirection.

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