Re: multi-threaded compiling

[Mischa Baars]

Hi Chet,

The only thing I can find about this behavior is the set +m / set -m
option, which disables / enables the notifications, but has no effect
otherwise.

The line from the script won't execute from the command line with either
option, neither enclosed with a 'bash -c' as in:

set +m; bash -c 'seconds=1; for (( i=0;i<32;i++ )); do exit ${i} & done;
sleep ${seconds}; for (( i=0;i<32;i++ )); do wait -np pid; e=${?}; echo
"$(printf %3u ${i}) pid ${pid} exit ${e}"; done;'

On Mon, Mar 11, 2024 at 8:20 PM Chet Ramey <chet.ramey@case.edu> wrote:

> On 3/11/24 2:50 PM, Mischa Baars wrote:
> > Which sort of brings us back to the original question I suppose. Who does
> > that line of code function from a script and why does it fail from the
> > command line?
>
> Job control and when the shell notifies the user about job completion,
> most likely, two of the relevant things that differ between interactive
> and non-interactive shells.
>
> --
> ``The lyf so short, the craft so long to lerne.'' - Chaucer
>                  ``Ars longa, vita brevis'' - Hippocrates
> Chet Ramey, UTech, CWRU    chet@case.edu    http://tiswww.cwru.edu/~chet/
>
>