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. (If,
on the other hand, I separate the declaration and the
definition of the variable as shown in the example below, the
$? variable is really set to the exit status of the external
tool.)
I realized this on version 4.1 (as this bug is reported for
now) as well as on older versions (4.0, 3.x), if I remember it
correctly.
Please check whether that behavior is intended or whether it's
really a bug.
Repeat-By:
The following script contains two functions, the first one
showing the scenario where the issue occurs (testfunc_fail),
the second one showing the working scenario (testfunc_ok).
testfunc_fail() {
local OUTPUT=$(touch /does/not/exist 2>&1)
echo "testfunc_fail() returned $? with message '$OUTPUT'."
}
testfunc_ok() {
local OUTPUT
OUTPUT=$(touch /does/not/exist 2>&1)
echo "testfunc_ok() returned $? with message '$OUTPUT'."
}
testfunc_fail
testfunc_ok
