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.