On 11/14/18 4:48 AM, Christopher Jefferson wrote:
>
> On 13/11/2018 14:59, Chet Ramey wrote:
>> On 11/13/18 4:28 AM, Christopher Jefferson wrote:
>>> Consider the following script. While the 3 sleeps are running, both jobs
>>> -p and $(jobs -p) will print 3 PIDs. Once the 3 children are finished,
>>> jobs -p will continue to print the 3 PIDs of the done Children, but
>>> $(jobs -p) will only print 1 PID. $(jobs -p) always seems to print at
>>> most 1 PID of a done child.
>> Since the $(jobs -p) is run in a subshell, its knowledge of its parent's
>> jobs is transient. In this case, the subshell deletes knowledge of the
>> jobs it inherits from its parent, but hangs onto the last asynchronous job
>> in case the subshell references $!.
>
>
> Is this a case of "works as intended" then? I find the current behaviour
> very strange -- I could understand if 'jobs -p' showed no information
> about processes spawned from the parent shell, or all of it, but the
> current position seems quite inconsistent.
Why? The shell has to keep the value and status of $! because POSIX says
you can always request it. A subshell inherits that pid, so it's `known'
to the subshell, and it will stick around until you get notfied of the
status. You don't restrict the `jobs -p' output to running jobs, so it's
going to show up, and you don't request the status of the job (even jobs
or jobs -l would be sufficient) so it's not going to go away.
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU c...@case.edu http://tiswww.cwru.edu/~chet/
