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Re: local failure

From: Laurent Picquet
Subject: Re: local failure
Date: Sat, 30 May 2020 15:33:20 +0100

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

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

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