Bash shell arguments

We frequently use arguments with the commands that we type on the command line interface. For example in the command $ls -l /tmp, ls is the command, l is the option & /tmp is the argument. We can use some arguments with our bash shell scripts as well to influence the behavior of the code as per our requirements.


$0
The first shell argument we look at is $0. This represents the name of the file/script which is being run. To illustrate I wrote a small script bargs.sh with the following content:

[root@centops ~]# cat bargs.sh
#!/bin/bash

echo "The name of the script is $0"

When I run the script, the name of the script gets substituted as the value of argument $0.

[root@centops ~]# bargs.sh

The name of the script is /root/bargs.sh

[root@centops ~]# ./bargs.sh

The name of the script is ./bargs.sh

Positional parameters

These arguments are the ones we type after the name of the script separated by spaces. They are $1, $2 & so on. Just a note, in case you are using more than nine arguments with the script then the 10th & subsequent arguments need to be represented by ${10} & so on, In case you write $10, the shell will interpret it as the value of $1 & append a zero to that value while printing.
I've modified the bargs.sh script used earlier to include the arguments $1 & $2.

[root@centops ~]# cat bargs.sh
#!/bin/bash

echo "The name of the script is $0"

echo "The first argument is $1"

echo "The second argument is $2"

I executed the script providing two arguments to it:

[root@centops ~]# ./bargs.sh sahil suri
The name of the script is ./bargs.sh
The first argument is sahil
The second argument is suri

$#
The argument $# represents the number of arguments typed with the script during execution. This is useful for applying a condition where in you want a script to run only when the user enters a required number of arguments. I updated the bargs.sh script as follows:

[root@centops ~]# cat bargs.sh
#!/bin/bash

echo "The name of the script is $0"

echo "The first argument is $1"

echo "The second argument is $2"

echo "you entered $# arguments"

The output from this script is as follows:

[root@centops ~]# ./bargs.sh sahil suri
The name of the script is ./bargs.sh
The first argument is sahil
The second argument is suri
you entered 2 arguments

$* and $@

Both these arguments basically perform the same function, They hold the arguments that were entered with the shell script. But there is a subtle difference in the interpretation. I've expanded the script bargs.sh further to illustrate their usage.

[root@centops ~]# cat bargs.sh

#!/bin/bash

echo "The name of the script is $0"

echo "The first argument is $1"

echo "The second argument is $2"

echo "you entered $# arguments"

echo "The arguments entered are $*"

echo "The arguments entered are $@"

The output of the script is as follows:

[root@centops ~]# ./bargs.sh sahil suri

The name of the script is ./bargs.sh

The first argument is sahil

The second argument is suri

you entered 2 arguments

The arguments entered are sahil suri

The arguments entered are sahil suri

The difference between $* & $@ isn't apparent from the above example. From what I've understood, the arguments $* and $@ will produce the same result unless we change the value of IFS (internal field separator) which is a white space by default. $* counts each argument as an individual string whereas $@ interprets all the arguments as a single string. 

With that understood, lets see an example to demonstrate this.

I've updated our bargs.sh script as follows:

[root@cclient1 ~]# cat bargs.sh

#!/bin/bash

echo -e "\e[34m illustrating bash arguments  \e[0m"

echo "the script name with path is:" $0

echo "the script name is:" `basename $0`

echo "the script location is:" `dirname $0`

echo "the 1st argument is:" $1

echo "the 2nd argument is:" $2

echo "the number of arguments are:" $#

echo "the arguments entered are" $*

echo "the arguments entered are" $@

echo "Changing field separator"

IFS='-'

echo "the arguments using \$*" "$*"

echo "the arguments using \$@" "$@"

The output from this script is:

[root@cclient1 ~]# ./bargs.sh sahil suri

        illustrating bash arguments

the script name with path is: ./bargs.sh

the script name is: bargs.sh

the script location is: .

the 1st argument is: sahil

the 2nd argument is: suri

the number of arguments are: 2

the arguments entered are sahil suri

the arguments entered are sahil suri

Changing field separator

the arguments using $* sahil-suri

the arguments using $@ sahil suri

In the above example I changed the value of IFS from the default white space to -. So $* interpreted sahil & suri as separate arguments & placed a - as separator between them but $@ did not.

$?
This isn't an argument supplied to a shell script. But I felt the need to mention this anyway in this article. The $? variable stores the exit status of the immediately executed command.An exit status of zero indicates successful execution of the program whereas a non-zero exit status indicates an unsuccessful execution. For example

[root@centops ~]# ls /
bin  boot  cgroup  check_dir  dev  etc  home  lib  lib64  lost+found  media  mnt  opt  Packages  proc  quadstor  R_D  root  sbin  selinux  srv  sys  tmp  usr  var
[root@centops ~]# echo $?
0

[root@centops ~]# ls /sahil

ls: cannot access /sahil: No such file or directory

[root@centops ~]# echo $?

2

This variable is critical to scripting because using it's value we can shape the flow of the script.