Special variables ($0, $1, $#, $@, $?, $) in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output of this script?

Consider this bash script saved as test.sh and run as ./test.sh apple banana:

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

What will be the output?

Bash Scripting

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

A

./test.sh
apple
3
apple banana
<process_id>

B

test.sh
banana
2
apple banana
<process_id>

C

./test.sh
apple
2
apple banana
<process_id>

D

./test.sh
banana
1
banana
<process_id>

💡 Hint

Remember $0 is the script name, $1 is the first argument, $# is the number of arguments, $@ is all arguments, and $$ is the current process ID.

$0 is the script name as called (./test.sh), $1 is the first argument (apple), $# counts arguments (2), $@ lists all arguments separated by spaces (apple banana), and $$ is the process ID (a number).

💻 Command Output

intermediate

What does $? represent after a command?

In bash, after running a command, what does the special variable $? hold?

Example:

ls /nonexistent
 echo $?

Bash Scripting

ls /nonexistent
 echo $?

A The name of the script

B The exit status of the last command (0 if success, non-zero if failure)

C The number of arguments passed to the script

D The process ID of the last command

💡 Hint

$? tells if the last command worked or not.

$? holds the exit status of the last command. Zero means success, any other number means an error or failure.

💻 Command Output

advanced

How many arguments are passed to this script?

Given this script run as ./script.sh one two three:

echo $#

What is the output?

Bash Scripting

echo $#

A 3

B 4

C 1

D 0

💡 Hint

$# counts how many arguments are given to the script.

$# counts the number of arguments passed, excluding the script name. Here, three arguments are passed.

💻 Command Output

advanced

What is the output of this script using $@ and $*?

Consider this script run as ./script.sh a b c:

echo "$@"
echo "$*"

What is the output?

Bash Scripting

echo "$@"
echo "$*"

A

a b c
"a b c"

B

"a" "b" "c"
a b c

C

"a b c"
a b c

D

a b c
a b c

💡 Hint

$@ and $* behave differently when quoted.

When quoted, "$@" expands each argument as a separate quoted string, so it prints each argument separated by spaces. "$*" expands all arguments as a single string separated by the first character of IFS (space by default), so it prints all arguments as one string.

💻 Command Output

expert

What is the value of $$ inside a script?

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

Example:

echo $$

Bash Scripting

echo $$

A The process ID of the current shell running the script

B The number of arguments passed to the script

C The exit status of the last command

D The name of the script

💡 Hint

$$ is related to the process running the script.

$$ holds the process ID (PID) of the current shell running the script. This is useful for creating unique temporary files or tracking the script process.

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