Re: Return status of command substitution with $(...) "gets lost"

[Chris F.A. Johnson]

On Thu, 4 Mar 2010, Ettelbrueck, Heiko wrote:

> Configuration Information [Automatically generated, do not change]:
> Machine: x86_64
> OS: linux-gnu
> Compiler: gcc
> Compilation CFLAGS:  -DPROGRAM='bash' -DCONF_HOSTTYPE='x86_64'
> -DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='x86_64-unknown-linux-gnu'
> -DCONF_VENDOR='unknown' -DLOCALEDIR='/usr/local/share/locale'
> -DPACKAGE='bash' -DSHELL -DHAVE_CONFIG_H   -I.  -I. -I./include -I./lib
>   -g -O2
> uname output: Linux wdfd00221495a 2.6.27.42-0.1-xen #1 SMP 2010-01-06
> 16:07:25 +0100 x86_64 x86_64 x86_64 GNU/Linux
> Machine Type: x86_64-unknown-linux-gnu
> 
> Bash Version: 4.1
> Patch Level: 0
> Release Status: release
> 
> Description:
>         There is a function which runs some external tool. Both the
>         tool's output and its exit status are relevant for further
>         execution of the bash script.
>         -> The tool's is stored into a (new) local variable, which is
>            then processed further.
>         -> The tool's exit status is retrieved from the $? variable.
> 
>         Important detail: The local variable is declared and defined in
>         the same step with "local VARNAME=$(do something)".
> 
>         Problem: The $? variable is always 0 after that statement.

   The return status is that of the command, local.

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