Special variables ($0, $1, $#, $@, $?, $) in Bash Scripting - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the special variable $0 represent in a bash script?

$0 holds the name of the script or command being executed. It helps you know which script is running.

beginner

What information does $1 provide in a bash script?

$1 is the first argument passed to the script. If you run ./script.sh hello, then $1 is hello.

beginner

Explain the purpose of $# in bash scripting.

$# tells you how many arguments were passed to the script. It helps check if the user gave the right number of inputs.

intermediate

What does $@ represent in a bash script?

$@ is all the arguments passed to the script, treated as separate words. Useful to loop over all inputs.

intermediate

What is the use of $? and $$ in bash?

$? shows the exit status of the last command (0 means success). $$ is the process ID of the current shell, useful for unique temp files.

In a bash script, what does $# represent?

A The number of arguments passed to the script

B The name of the script

C The first argument

D The process ID of the script

$# counts how many arguments you gave to the script.

Which special variable holds the exit status of the last command?

A $0

B $?

C $@

D $$

$? tells if the last command worked (0) or failed (non-zero).

If you want to access all arguments passed to a bash script as separate words, which variable do you use?

A $1

B $0

C $#

D $@

$@ holds all arguments separately, perfect for loops.

What does $$ represent in a bash script?

A The process ID of the current script

B The script name

C The first argument

D The number of arguments

$$ is the unique ID of the running script process.

What value does $0 hold inside a bash script?

A The last argument

B The number of arguments

C The script's filename or command name

D The exit status of last command

$0 is the name of the script or command you ran.

Describe the purpose of each special variable: $0, $1, $#, $@, $?, and $$ in bash scripting.

Think about what each variable tells you about the script or its inputs.

How would you use special variables to check if a user passed exactly two arguments to your bash script?

Counting arguments helps control script behavior.

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