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

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From Oğuz <oguzismailuysal@gmail.com>
Newsgroups gnu.bash.bug
Subject Re: 'foo > >(bar)' doesn't set $! for external foo not invoked via 'command'
Date Mon, 20 Jul 2020 08:10:06 +0300
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19 Temmuz 2020 Pazar tarihinde Rusty Bird <rustybird@net-c.com> yazdı:

> Configuration Information [Automatically generated, do not change]:
> Machine: x86_64
> OS: linux-gnu
> Compiler: gcc
> Compilation CFLAGS: -O2 -g -pipe -Wall -Werror=format-security
> -Wp,-D_FORTIFY_SOURCE=2 -Wp,-D_GLIBCXX_ASSERTIONS -fexceptions
> -fstack-protector-strong -grecord-gcc-switches -specs=/usr/lib/rpm/redhat/redhat-hardened-cc1
> -specs=/usr/lib/rpm/redhat/redhat-annobin-cc1 -m64 -mtune=generic
> -fasynchronous-unwind-tables -fstack-clash-protection -fcf-protection
> -Wno-parentheses -Wno-format-security
> uname output: Linux [...] 4.19.132-1.pvops.qubes.x86_64 #1 SMP Tue Jul 14
> 03:42:21 UTC 2020 x86_64 x86_64 x86_64 GNU/Linux
> Machine Type: x86_64-redhat-linux-gnu
>
> Bash Version: 5.0
> Patch Level: 17
> Release Status: release
>
> Description:
>         $! isn't set to the PID of 'bar' for
>
>                 foo > >(bar)
>
>         if 'foo' is an external program (as opposed to a builtin or
>         function) - unless it's invoked via 'command'.
>
> Repeat-By:
>         type date  # "date is /usr/bin/date"
>         set -u
>         echo $!  # "bash: $!: unbound variable"
>
>         date > >(sleep 20)
>         echo $!  # "bash: $!: unbound variable"
>
>         command date > >(sleep 21)
>         echo $!  # "123"
>
>         date_wrapper() { date; }
>         date_wrapper > >(sleep 22)
>         echo $!  # "234"
>
>
> 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.


> This only works if 'program' is changed to 'command program'.
>
> Rusty
>


-- 
Oğuz

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Re: 'foo > >(bar)' doesn't set $! for external foo not invoked via 'command' Oğuz <oguzismailuysal@gmail.com> - 2020-07-20 08:10 +0300

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