Re: wait -n misses signaled subprocess

[Chet Ramey]

On 1/29/24 12:07 PM, Chet Ramey wrote:
> On 1/29/24 7:12 AM, Robert Elz wrote:
> >      Date:        Sun, 28 Jan 2024 18:21:42 -0500
> >      From:        Chet Ramey <chet.ramey@case.edu>
> >      Message-ID:  <3347f790-529b-4bee-91fd-de39bed3f49e@case.edu>
> >
> >    | because `wait -n' doesn't look in the table
> >    | of saved statuses -- its job is to wait for `new' jobs to terminate, 
> > not
> >    | ones that have already been removed from the table.
> >
> > That's very interesting, and most unexpected information.
> >
> > I always wondered why the option was 'n' - I would have made it
> > be 'a' probably, as a shorthand for "any" - but then I decided
> > that perhaps 'n' was a better choice, as "a" could also be "all",
> > the option name would not be providing any real clue at all, so
> > I assumed you'd been ultra clever and used 'n' as the next char
> > in "any" and also as it can be read like the first part of "en" "ee"
> > (which you need to say out loud, or at least in your head, to get the
> > effect of).
>
> You should have. You told me about your implementation using `-n' in
> 10/2017, long before I implemented it (4/2020).

Sorry, this is my mistake. That was a different feature. Bash implemented
`wait -n' first.

--
``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/

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