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Re: Cannot grep jobs list when jobs called in a function


From: Reuti
Subject: Re: Cannot grep jobs list when jobs called in a function
Date: Fri, 29 Apr 2016 15:32:22 +0200

> Am 29.04.2016 um 14:15 schrieb Greg Wooledge <address@hidden>:
> 
> On Thu, Apr 28, 2016 at 10:38:53AM -0600, Eric Blake wrote:
>> Bash has code to special-case 'jobs |' when it can obviously tell that
>> you are running the jobs builtin as the sole command of the left side of
>> a pipe, to instead report about the jobs of the parent shell,
> 
> Oh, that's interesting.  I didn't know that.

Me too.

I understand that the behavior of the builtin `jobs` changes, in case it 
discovers that it's run a subshell. But what is happening here:

$ sleep 300 &
[1] 31766
$ function lister() { date; jobs; }
$ lister
Fri Apr 29 15:29:46 CEST 2016
[1]+  Running                 sleep 300 &
$ lister | cat
Fri Apr 29 15:30:00 CEST 2016
[1]   Done                    date

My question is: why does the `date` command show up as "done" at all? I would 
expect the output to be just empty.


> but that
>> special-case code cannot kick in if you hide the execution of jobs,
>> whether by hiding it inside a function as you did, or by other means
>> such as:
>> $ eval jobs | grep vim
> 
> In general, if you want to filter the output of "jobs" or some other
> builtin that changes its behavior when invoked in a subshell, then
> you need to avoid the subshell.  That means no pipeline, no command
> substitution, etc.  Basically that leaves you with a temporary file.
> 
> tmpfile=...  # boilerplate code to create a temp file on whatever OS
> trap 'rm -f "$tmpfile"' EXIT
> jobs > "$tmpfile"
> if grep -q vim "$tmpfile"; then ...

Depending on the overall program, this might work to avoid a subshell:

if grep -q vim < <(realjobs); then ...

-- Reuti





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