Re: why must non-standard $IFS members be treated so differently ?

[Chris F.A. Johnson]

On Sun, 29 Jul 2012, Jason Vas Dias wrote:

> Good day Chet, list -
> I'm concerned about the difference in output of these functions with
> the example input
> given on the '$' prefixed line below (with 4.2.29(2)-release
> (x86_64-unknown-linux-gnu)):
>
>     function count_args     {                v=($@);  echo address@hidden; }

   Always quote address@hidden Without quotes, it's the same as $*

function count_args     {                v=( "$@" );  echo address@hidden; }

--
   Chris F.A. Johnson, <http://cfajohnson.com/>
   Author:
   Pro Bash Programming: Scripting the GNU/Linux Shell (2009, Apress)
   Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)