GNU bug report logs - #62406
“! failing-command” pattern in shell tests is wrong

Previous Next

Full log

View this message in rfc822 format

From: Martin Castillo <castilma <at> uni-bremen.de>
To: Ludovic Courtès <ludo <at> gnu.org>, 62406 <at> debbugs.gnu.org
Cc: Eric Bavier <bavier <at> posteo.net>
Subject: bug#62406: “! failing-command” pattern in shell tests is wrong
Date: Sun, 26 Mar 2023 20:16:38 +0200

Hi,

Am 23.03.23 um 17:00 schrieb Ludovic Courtès:> In 
<https://issues.guix.gnu.org/43616> (commit
> d8934360d2453a403b5433e71d09188e4ed23b57), we changed:
>
>    if command that should fail; then false; else true; fi
>
> to:
>
>    ! command that should fail
>
> I had reservations back then, and now I know why: :-)
>
> --8<---------------cut here---------------start------------->8---
> $ bash -xe -c '! true; true'
> + true
> + true
> $ echo $?
> 0
> $ bash -xe -c '! false; true'
> + false
> + true
> $ echo $?
> 0
> --8<---------------cut here---------------end--------------->8---
>
> Whether or not the command following the exclamation mark succeeds, the
> statement succeeds.  Bummer.
>
> The Bash manual (info "(bash) Pipelines") reads:
>
>    If the reserved word '!' precedes the pipeline, the exit status is the
>    logical negation of the exit status as described above.  The shell
>    waits for all commands in the pipeline to terminate before returning a
>    value.
>
> To me, that means it should work as we thought, but it’s a fact that it
> doesn’t.
the documentation on `-e` says:

-e      Exit  immediately  if a pipeline (which may consist of a
        single simple command), a list, or  a  compound  command
        (see SHELL GRAMMAR above), exits with a non-zero status.
        The shell does not exit [...] if the
        command's return value is being inverted with !.

I have no idea what might be the rationale for that.

Previous Next

GNU bug tracking system
Copyright (C) 1999 Darren O. Benham, 1997,2003 nCipher Corporation Ltd, 1994-97 Ian Jackson.