"jobs" builtin: print zero exit status

[Dmitry Bolshakov]

   Hi!
   imho "jobs" should either print zero exit status or it should be
   documented that "Done" means zero exit status
   $ bash -c 'set -m; perl -e "sleep 1; exit 123"& { sleep 3; jobs -l; }'
   [1]+е║ Exit 123е║е║е║е║е║е║е║е║е║е║е║е║е║е║е║ perl -e "sleep 1; exit 123"
   ok, I see exit status
   but
   $ bash -c 'set -m; perl -e "sleep 1; exit 0"& { sleep 3; jobs -l; }'
   [1]+е║ Doneе║е║е║е║е║е║е║е║е║е║е║е║е║е║е║е║е║е║е║ perl -e "sleep 1; exit 0"
   imho "Exit 0" would be better for scripting

   --

   with best regards
   Dmitry Bolshakov

Re: "jobs" builtin: print zero exit status

[Chet Ramey]

On 9/12/13 6:14 AM, Dmitry Bolshakov wrote:
>Hi!
>imho "jobs" should either print zero exit status or it should be
>documented that "Done" means zero exit status
>$ bash -c 'set -m; perl -e "sleep 1; exit 123"& { sleep 3; jobs -l; }'
>[1]+е║ Exit 123е║е║е║е║е║е║е║е║е║е║е║е║е║е║е║ perl -e "sleep 1; exit 123"
>ok, I see exit status
>but
>$ bash -c 'set -m; perl -e "sleep 1; exit 0"& { sleep 3; jobs -l; }'
>[1]+е║ Doneе║е║е║е║е║е║е║е║е║е║е║е║е║е║е║е║е║е║е║ perl -e "sleep 1; exit 0"
>imho "Exit 0" would be better for scripting

The format of the text isn't in the man page at all.  Parsing the output
of jobs is a really terrible way to figure out a job's exit status.  You
might try using `wait'.

Chet
-- 
``The lyf so short, the craft so long to lerne.'' - Chaucer
 ``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRUc...@case.eduhttp://cnswww.cns.cwru.edu/~chet/