> >>> 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,
[...]
... hey, I think I just figured out the GOAL!
You want to run a whole bunch of jobs in parallel, but only 3 at a
time. Right? There
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,
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
On Tue, Nov 13, 2018 at 09:59:51AM -0500, 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
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
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
