Need help in explaining Bash code

Linh · 07-08-2004, 11:07 AM

The code below is stored in a file named
/home/bash_script

1) When I run the program bash_script, it did not did
list all files and directory in argument two ($2) which is
/usr/suid, but instead it list the files and directory in
/home/

2) On the code abc /usr/suid "$2", I change it to
abc /usr/suid "$1", but it did not changed anything.

3) The code abc /usr/suid "$2" . Does it means to
call the function abc whereas usr/suid is passed as
a second argument ?
========================

Code:

#!/bin/bash

function abc()
{
   ls -al $2
}

abc /usr/suid  "$2"

SheldonPlankton · 07-08-2004, 11:24 AM

I think you want ...

Code:

#!/bin/bash
function abc() {
ls -al $1
}

abc /usr/suid

Linh · 07-08-2004, 01:03 PM

Hi SheldonPlankton, and thank you for your help.
What does the statement abc /usr/suid "$2" do ?
Does it passed an external argument number two to the function abc ?

function abc()
{
ls -al $2
}

abc /usr/suid "$2"

Komakino · 07-08-2004, 07:18 PM

Yeah, but as you have no argument 2 it would just call 'ls -al' (hence it's operating on the current directory)

jlliagre · 07-09-2004, 10:50 AM

Well, you may have an argument 2 if you pass it to the original script:

$ bash_script unused_arg1 arg2

That would lead to execute:

ls -al arg2