Re: [Help-bash] "local" modifies the behavior of "$?"

[Dan Douglas]

On Monday, June 24, 2013, Chet Ramey <address@hidden> wrote:
> On 6/23/13 11:39 PM, Dan Douglas wrote:
>
>> Not documented but frequently asked.
>
> Not quite.  From the `SIMPLE COMMAND EXPANSION' section of the man page:
>
> "If  there is a command name left after expansion, execution proceeds as
> described below.  Otherwise, the command exits.  If one of  the  expan-
> sions  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."
>
> Similarly, the descriptions of `local' and `declare' list the conditions
> under which they will return non-zero.
>
> Chet
>
> --
> ``The lyf so short, the craft so long to lerne.'' - Chaucer
>                  ``Ars longa, vita brevis'' - Hippocrates
> Chet Ramey, ITS, CWRU    address@hidden    http://cnswww.cns.cwru.edu/~chet/
>
Yup, what Chet said... You guys might also be interested in this table I put together here: http://wiki.bash-hackers.org/dict/terms/exit_status#portability. It lists a good number of corner-cases like "! return" and others.