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Re: multi-threaded compiling


From: Mischa Baars
Subject: Re: multi-threaded compiling
Date: Tue, 12 Mar 2024 13:32:45 +0100

On Tue, Mar 12, 2024 at 12:30 PM Greg Wooledge <greg@wooledge.org> wrote:

> On Tue, Mar 12, 2024 at 02:15:38PM +0700, Robert Elz wrote:
> >     Date:        Mon, 11 Mar 2024 17:25:57 -0400
> >     From:        Chet Ramey <chet.ramey@case.edu>
> >     Message-ID:  <322e10a6-3808-49be-aa9d-a1d367a90572@case.edu>
> >
> >   | OK, here's the longer answer. When the shell is interactive, and job
> >   | control is enabled, the shell prints job completion notifications to
> >   | stdout at command boundaries.
> >
> > which is, IMO, yet another bogus bash misfeature.  That should
> > happen, with the effect described, but only when PS1 is about
> > to be printed - precisely so commands like the one described
> > will work.   Sigh.
> >
> > There aren't many complaints about this misfeature of bash,
> > as almost no-one writes interactive command lines where it makes
> > a difference.   That doesn't mean they should not be able to.
>
> Yeah, it appears you're right.  In a script, this works as expected:
>
> hobbit:~$ cat foo
> #!/bin/bash
> for i in {0..3}; do sleep 1 & done
> for i in {0..3}; do
>     wait -n -p pid; e=$?
>     printf 'pid %s status %s\n' "$pid" "$e"
> done
> hobbit:~$ ./foo
> pid 530359 status 0
> pid 530360 status 0
> pid 530361 status 0
> pid 530362 status 0
>
>
> But interactively, *even with bash -c and set +m*, it just fails:
>
> hobbit:~$ bash -c 'set +m; for i in {0..3}; do sleep 1 & done; for i in
> {0..3}; do wait -n -p pid; e=$?; printf "pid %s status %s\n" "$pid" "$e";
> done'
> pid 530407 status 0
> pid 530410 status 0
> pid  status 127
> pid  status 127
>
> Looks like a race condition, where some of the children get reaped and
> tossed away before "wait -n -p pid" has a chance to grab their status.
>
> If I stick a "sleep 2" in between the two loops, then it's even worse:
>
> hobbit:~$ bash -c 'set +m; for i in {0..3}; do sleep 1 & done; sleep 2;
> for i in {0..3}; do wait -n -p pid; e=$?; printf "pid %s status %s\n"
> "$pid" "$e"; done'pid  status 127
> pid  status 127
> pid  status 127
> pid  status 127
>
> ALL of the children are discarded before the second loop has a chance to
> catch a single one of them. * This is clearly not working as expected*.
>

Now we're talking :) To observe different output from script and from
Makefile certainly is somewhat peculiar and unexpected.


>
> Using "bash +i -c" doesn't change anything, either.  Or "bash +i +m -c".
> Whatever magic it is that you get by putting the code in an actual
> script, I can't figure out how to replicate it from a prompt.
>
>


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