From: Peter Stephenson <p.w.stephenson@ntlworld.com>
To: zsh-workers@zsh.org
Subject: Re: set -e (no && or ||)
Date: Fri, 12 Oct 2012 12:53:35 +0100 [thread overview]
Message-ID: <CAECNH1RQxAH3DvSCVagR_DQHqA6Cpi2U+0p9m3=rhp=kDSnGCQ@mail.gmail.com> (raw)
In-Reply-To: <20121012110542.314890@gmx.com>
On Fri, 12 Oct 2012 07:05:41 -0400
Sergey Fadeev <hindsight@email.com> wrote:
> Why doesn't it exit the shell?
> $ set -e
> $ echo $(false)
> Shouldn't the error code of $(false) command substitution be checked
> by set -e before passing stdout to the echo builtin?
No, because the command was "echo", and that didn't fail. Exit status
effectively means exit status seen by the main shell command
interpreter ($?), although I'm sure there are some subtleties I haven't
thought about.
The way to get the status of a substitution to fail is to use an
assignment:
output=$(false)
which does cause the shell to exit on failure, because it would set $? to
1. This is standard shell behaviour, though I can't point to where in
the standard it says.
pws
next prev parent reply other threads:[~2012-10-12 12:00 UTC|newest]
Thread overview: 4+ messages / expand[flat|nested] mbox.gz Atom feed top
2012-10-12 11:05 Sergey Fadeev
2012-10-12 11:53 ` Peter Stephenson [this message]
2012-10-12 13:29 ` Bart Schaefer
2012-10-12 13:49 ` Chet Ramey
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