Re: local failure

[Oğuz]

30 Mayıs 2020 Cumartesi tarihinde Laurent Picquet <lpicquet@gmail.com>
yazdı:

> Hello Dale,
>
> This is really interesting.
> Should the 'local' command be the one able to detect that the assignment to
> the variable had an non-zero exit code and return the non-zero exit code?
>
> as a developer, it is counter-intuitive that the 'local' command tells us
> everything is ok when it wasn't. If feel it should know that the assignment
> encountered a problem and should report it
>
>
Everything is ok for `local` though; it takes a valid assignment statement
and successfully evaluates that. So it's not that the assignment
encountered a problem, but that the expansion has failed, which has nothing
to do with `local`. So there is no reason for `local` to return a non-zero
exit status in that case.


> The return status is zero unless local is used outside a function, an
> invalid name is supplied, or name is a readonly variable.
>
>
>
> On Fri, 29 May 2020 at 03:43, Dale R. Worley <worley@alum.mit.edu> wrote:
>
> > It's a subtle point.  See this paragraph in the bash manual page:
> >
> >        If there is a command name left after expansion, execution
> >        proceeds as described below.  Otherwise, the command exits.  If
> >        one of the expansions contained a command substitution, the exit
> >        status of the command is the exit status of the last command
> >        substitution performed.  If there were no command substitutions,
> >        the command exits with a status of zero.
> >
> > In one of your examples, a "local" command is generated using a command
> > substitution, so the exit status is that of the local command.  In the
> > other, only an assignment is done, which is not a command, so the exit
> > status is that of the last command substitution.
> >
> > Dale
> >
>
>
> --
>
>
> --
>
> Laurent Picquet
>
> 16, Hunters Chase
>
> South Godstone
>
> RH98HR
>
> England
>
> tel: 07882 356 104
>


-- 
Oğuz