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Re: value of $? and exit status are partially undocumented, possibly wro
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From: |
Chet Ramey |
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Subject: |
Re: value of $? and exit status are partially undocumented, possibly wrong |
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Date: |
Tue, 17 Oct 2023 10:16:35 -0400 |
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User-agent: |
Mozilla Thunderbird |
On 10/13/23 9:18 AM, Vincent Lefevre wrote:
With bash 5.2.15(1)-release under Debian (bash 5.2.15-2+b6 package),
on one of my machines:
$ bash -c 'echo $?'
0
$ bash
vlefevre@cventin:~$ echo $?
0
$ bash ; echo $?
vlefevre@cventin:~$ exit
exit
0
and on another one:
$ bash -c 'echo $?'
0
$ bash
vinc17@zira:~$ echo $?
1
$ bash ; echo $?
vinc17@zira:~$ exit
exit
1
The difference seems to be due to the exit status of the last command
of the .bashrc file. But this is undocumented.
How so? The exit status ($?) is the status of "the last command executed,"
and an interactive non-login shell "reads and executes commands from
~/.bashrc."
Note that on the opposite, for a login shell, e.g. "bash -l", the
exit status of the last command of .bash_logout is ignored for the
exit status of bash. So this is confusing.
Think of it like the EXIT trap. If you run `exit 4', you don't want
~/.bash_logout changing that exit status.
Moreover, for "bash -l /dev/null", when .bash_profile ends with a
non-zero exit status, the behavior contradicts
Commands are executed. They are read and executed from .bash_profile.
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU chet@case.edu http://tiswww.cwru.edu/~chet/