Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > gnu.bash.bug > #16588

Re: 'foo > >(bar)' doesn't set $! for external foo not invoked via 'command'

Show all headers | View raw

[Multipart message — attachments visible in raw view] - view raw

Oğuz:
> > For context - I'm filtering a program's stdout and stderr
> > (separately), requiring successful exit statuses for the program and
> > both filters:
> >
> >         set -u -o pipefail
> >         { program 2> >(stderr_filter >&2) && wait $!; } | stdout_filter &&
> > ...

> Not sure if process substitution is really necessary here,
> 
>     set -u -o pipefail
>     { program 2>&3 | stdout_filter;  } 3>&1 | stderr_filter && ...
> 
> does the same thing.

That one filters program's stdout twice - first through stdout_filter
and then through stderr_filter - with program's stdout and stderr both
finally arriving at stdout. But tweaked like this, it seems to cover
all the bases:

        set -u -o pipefail
	{
	    program 2>&1  >&"$out" {out}>&- |
	    stderr_filter >&2      {out}>&-
	} {out}>&1 | stdout_filter

And it even waits for stderr_filter if program failed. My original
snippet neglected that case, otherwise it would have looked more like

       set -u -o pipefail
       (
           trap 'wait $! || exit $?' EXIT
           command program 2> >(stderr_filter >&2)
       ) | stdout_filter

which isn't exactly pretty either, even if the bug(?) requiring
'command' is fixed.

Rusty

Back to gnu.bash.bug | Previous | Next | Find similar

Thread

Re: 'foo > >(bar)' doesn't set $! for external foo not invoked via 'command' Rusty Bird <rustybird@net-c.com> - 2020-07-20 10:21 +0000

csiph-web