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)
