Special variables ($0, $1, $#, $@, $?, $) in Bash Scripting - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print the name of the script being run.

Bash Scripting

echo "Script name is: [1]"

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

A$0

B$1

C$$

D$#

Attempts:

3 left

💡 Hint

Common Mistakes

Using $1 which is the first argument, not the script name.

Using $$ which is the process ID, not the script name.

The special variable $0 holds the name of the script being executed.

2fill in blank

medium

Complete the code to print the number of arguments passed to the script.

Bash Scripting

echo "Number of arguments: [1]"

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

A$0

B$@

C$#

D$?

Attempts:

3 left

💡 Hint

Common Mistakes

Using $@ which lists all arguments, not the count.

Using $? which shows the last command's exit status.

The special variable $# holds the count of arguments passed to the script.

3fill in blank

hard

Fix the error in the code to print all arguments as separate words.

Bash Scripting

echo [1]

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

A$#

B$0

C$?

D$@

Attempts:

3 left

💡 Hint

Common Mistakes

Using $# which gives only the number of arguments.

Using $0 which is the script name.

The special variable $@ expands to all arguments as separate words.

4fill in blank

hard

Fill both blanks to check if the last command succeeded and print its exit code.

Bash Scripting

if [ [1] -eq 0 ]; then
  echo "Success with exit code [2]"
fi

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

A$?

B$#

C$0

D$$

Attempts:

3 left

💡 Hint

Common Mistakes

Using $$ which is the process ID, not exit status.

Using $# which counts arguments, not exit status.

The special variable $? holds the exit status of the last command.

5fill in blank

hard

Fill all three blanks to print the script name, process ID, and all arguments.

Bash Scripting

echo "Script: [1]"
echo "PID: [2]"
echo "Args: [3]"

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

A$0

B$$

C$@

D$#

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up $@ and $# for arguments.

Using $? instead of $$ for process ID.

Use $0 for script name, $$ for process ID, and $@ for all arguments.

Practice

1. In a bash script, what does the special variable $# represent?

easy

A. The exit status of the last command

B. The name of the script

C. The number of arguments passed to the script

D. The process ID of the script

Solution

1. Step 1: Understand the role of $#

   The variable $# in bash holds the count of all arguments passed to the script.

2. Step 2: Differentiate from other special variables

   $0 is the script name, $$ is the process ID, and $? is the last command's exit status, so they don't represent argument count.

3. Final Answer:

   The number of arguments passed to the script -> Option C

4. Quick Check:

   $# = argument count [OK]

Hint: Remember: '#' means count, so $# counts arguments [OK]

Common Mistakes:

• Confusing $# with $@ which holds all arguments
• Thinking $0 is the argument count
• Mixing up $? with argument count

2. Which of the following is the correct way to print the first argument passed to a bash script?

easy

A. echo $1

B. echo $0

C. echo $#

D. echo $$

Solution

1. Step 1: Identify the variable for the first argument

   $1 holds the first argument passed to the script.

2. Step 2: Check other options

   $0 is the script name, $# is the argument count, and $$ is the script's process ID, so they are incorrect for printing the first argument.

3. Final Answer:

   echo $1 -> Option A

4. Quick Check:

   First argument = $1 [OK]

Hint: Use $1 to access the first argument in bash scripts [OK]

Common Mistakes:

• Using $0 instead of $1 for first argument
• Confusing $# with argument value
• Trying to print $$ for argument

3. Consider the script ./myscript.sh apple banana with this content:

echo "$0"
echo "$1"
echo "$#"
echo "$@"
echo "$$"

What will be the output of the third line echo "$#"?

medium

A. myscript.sh

B. 3

C. apple

D. 2

Solution

1. Step 1: Count the arguments passed

   The script is called with two arguments: 'apple' and 'banana'. So, the count is 2.

2. Step 2: Understand what $# prints

   $# prints the number of arguments, which is 2 here.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Arguments count = 2 [OK]

Hint: Count arguments after script name for $# [OK]

Common Mistakes:

• Counting script name as argument
• Confusing $@ with $#
• Expecting argument values instead of count

4. You wrote this script snippet:

echo "Last command status: $?"
ls /nonexistent

But it always prints Last command status: 0 even if the directory doesn't exist. What is the problem?

medium

A. The ls command is correct, no problem

B. The echo runs before ls, so $? is from previous command

C. You should use $$ instead of $?

D. You need to quote $? differently

Solution

1. Step 1: Understand when $? is evaluated

   $? holds the exit status of the last command executed before it.

2. Step 2: Check command order

   The echo runs first, so it shows the status of the command before echo, not ls.

3. Final Answer:

   The echo runs before ls, so $? is from previous command -> Option B

4. Quick Check:

   Check command order for $? [OK]

Hint: Print $? immediately after the command to get correct status [OK]

Common Mistakes:

• Using $$ instead of $? for exit status
• Printing $? before running the command
• Assuming $? holds script PID

5. You want to write a bash script that prints its own process ID and all arguments passed to it on separate lines. Which script snippet correctly does this?

hard

A. echo "PID: $$"
echo "Args: $@"

B. echo "PID: $0"
echo "Args: $#"

C. echo "PID: $?"
echo "Args: $1"

D. echo "PID: $1"
echo "Args: $$"

Solution

1. Step 1: Identify variable for process ID

   $$ holds the current script's process ID.

2. Step 2: Identify variable for all arguments

   $@ holds all arguments passed to the script as separate words.

3. Final Answer:

   echo "PID: $$" echo "Args: $@" -> Option A

4. Quick Check:

   PID = $$, all args = $@ [OK]

Hint: Use $$ for PID and $@ for all arguments [OK]

Common Mistakes:

• Using $0 as PID instead of $$
• Using $# (count) instead of $@ (all args)
• Mixing $1 (first arg) with PID