Oh, sorry. Without the -n that is.
bash -c 'set +m; seconds=1; for (( i=0;i<32;i++ )); do exit ${i} & done;
sleep ${seconds}; for (( i=0;i<32;i++ )); do wait -p pid; e=${?}; echo
"$(printf %3u ${i}) pid ${pid} exit ${e}"; done;'
bash -c 'set -m; seconds=1; for (( i=0;i<32;i++ )); do exit ${i} & done;
sleep ${seconds}; for (( i=0;i<32;i++ )); do wait -p pid; e=${?}; echo
"$(printf %3u ${i}) pid ${pid} exit ${e}"; done;'
disables / enables the notifications respectively, but doesn't do anything
otherwise.
On Mon, Mar 11, 2024 at 8:20 PM Chet Ramey <chet.ra...@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 c...@case.edu http://tiswww.cwru.edu/~chet/
>
>
