Shell Programming and Scripting

I'm writing a wrapper script (in bash) that wraps another (bash) script. When calling the wrapped script, I need to pass all the received arguments/options to it. Is there a built in variable that holds all the options? I wrote a little while loop (see below) which works. But I wanted to know if there was an easier way.

Thanks.

What I have now:

Code:

#!/bin/bash

#Code that sets some environment variables

ARGS=""
while [ "$1" != "" ]; do
    ARGS="${ARGS} $1"
    shift
done

. $(dirname ${0})/startup.sh ${ARGS}

---------- Post updated at 02:36 PM ---------- Previous update was at 01:45 PM ----------

I found that I can use $@ (I just had trouble finding a good Google search query)

Thanks.

For other people's future reference, see: Reference Cards

Internal Variables

$*
All of the positional parameters, seen as a single word
Note "$*" must be quoted.
$@
Same as $*, but each parameter is a quoted string,
that is, the parameters are passed on intact, without interpretation or expansion.
This means, among other things, that each parameter in the argument list is seen as a separate word.
Note Of course, "$@" should be quoted.