On Wed, Apr 9, 2014 at 3:16 PM, Håkon Bugge <hakon.bu...@gmail.com> wrote:
> > On 9. apr. 2014, at 14.04, Greg Wooledge wrote: > > > On Wed, Apr 09, 2014 at 12:43:40PM +0200, Håkon Bugge wrote: > >> This script never terminates: > >> -------------------------------------------------- > >> #!/bin/bash > >> > >> for P in `seq 5`; do > >> sleep 1& > >> done > >> > >> while true; do > >> usleep 20000 > >> set foo `jobs -p` > >> LEFT=$# > >> LEFT=$[LEFT-1] > >> echo $LEFT jobs left > >> if [ x$LEFT = x0 ]; then > >> break > >> fi > >> done > > > >> Pasting the same commands in an interactive shell, its works. > > > > Interactive shells enable job control (monitor mode), whereas > > noninteractive shells (scripts) do not. If you want to use job control > > commands (like "jobs") within a script, you must enable monitor mode > > (set -m, or set -o monitor, or #!/bin/bash -m). > > That is not the issue. Try it out. > > > Håkon > I think greg hints that what "jobs" should do without set -m is not defined ...I wonder if it's really the case However enabling set -m doesn't help I agree that your script triggers a strange behavior: Funnily : bash -c 'sleep 1& sleep 1& sleep 1& sleep 3; jobs -p' # doesn't print anything while if you add just "jobs -p" in your loop without capturing it always print all the pids
