Special variables ($0, $1, $#, $@, $?, $) in Bash Scripting

Introduction

Special variables help you get information about your script and its inputs easily.

When you want to know the name of the script being run.

When you need to access the arguments passed to your script.

When you want to count how many arguments were given.

When you want to check if the last command worked or failed.

When you want to know the process ID of your running script.

Syntax

Bash Scripting

$0, $1, $#, $@, $?, $$

$0 is the script name.

$1 is the first argument, $2 the second, and so on.

$# is the number of arguments.

$@ is all arguments as separate words.

$? is the exit status of the last command (0 means success).

$$ is the process ID of the current script.

Examples

Shows script name, first argument, and how many arguments were passed.

Bash Scripting

echo "Script name: $0"
echo "First argument: $1"
echo "Number of arguments: $#"

Prints all arguments separated by spaces.

Bash Scripting

echo "All arguments: $@"

Runs a command that fails, then shows its exit status.

Bash Scripting

ls /notexist
 echo "Last command exit status: $?"

Prints the current script's process ID.

Bash Scripting

echo "My process ID is $$"

Sample Program

This script prints the script name, all arguments, number of arguments, first argument, runs 'ls /tmp', shows if it succeeded, and prints its own process ID.

Bash Scripting

#!/bin/bash

# Show script name
 echo "Script name: $0"

# Show all arguments
 echo "Arguments: $@"

# Show number of arguments
 echo "Number of arguments: $#"

# Show first argument
 echo "First argument: $1"

# Run a command
 ls /tmp

# Show exit status of last command
 echo "Exit status of ls: $?"

# Show process ID
 echo "Process ID: $$"

Important Notes

If you run the sample script with: ./script.sh arg1 arg2, $1 is 'arg1' and $# is 2.

Exit status ($?) is 0 if the last command worked, any other number means error.

Process ID ($$) is useful if you want to create temporary files unique to your script.

Summary

Special variables give quick info about your script and its inputs.

$0 is script name, $1 is first argument, $# counts arguments.

$@ holds all arguments, $? shows last command success, $$ is script's process ID.

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