How to pass all arguments to a function

stormcloud · 07-06-2010, 10:24 AM

I've got a sticky little problem with a bash script. Please consider the following code:

Code:

  #! /bin/bash

  processArgs() {
    echo "Count = $#"

    while [[ $# -ne 0 ]]; do
      echo $1
      shift
    done
  }

  processArgs $*

If I call this script with

Code:

  ./script first second third

it'll print each of the argument on a new line - exactly what I would expect. However if I call it with

Code:

  ./script "Single Argument" "Second-Argument"

it splits the first argument in two using the space as a delimitor.

The problem appears to be the call to processArgs, where $* doesn't honour the quotes around the variables sent to the script.

Any ideas how to get around this?

zirias · 07-06-2010, 10:33 AM

AFAIK, the notation "$@" (including the quotes) should produce a correctly re-quoted list of the arguments like they were initially passed. So you could try this instead of $*.

stormcloud · 07-07-2010, 02:38 AM

That works :-)

I had tried plain old $@ (without quotes), which didn't work. I didn't realise that I needed the quotes.

Thanks.

zirias · 07-07-2010, 05:32 AM

The special thing about $@ is that, when found inside double quotes, it results in EACH of the positional arguments double-quoted individually. $* in double quotes will instead double-quote the whole argument list, resulting in only one big string.

As far as I remember, there is no difference between them when not quoted (just replaced by the plain argument list, thus letting the shell split it at whitespace characters)

stormcloud · 07-08-2010, 03:15 AM

Ah! I was wondering what the difference between them was.

Cheers