coproc state wrongly inherited by subshell

[Tobias Hoffmann]

Configuration Information [Automatically generated, do not change]:
Machine: i586
OS: linux-gnu
Compiler: gcc
Compilation CFLAGS: -DPROGRAM='bash' -DCONF_HOSTTYPE='i586' 
-DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='i586-pc-linux-gnu' 
-DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash' 
-DSHELL -DHAVE_CONFIG_H -I. -I../. -I.././include -I.././lib 
-D_FORTIFY_SOURCE=2 -g -O2 -fstack-protector-strong -Wformat 
-Werror=format-security -Wall
uname output: Linux worker 4.13.0-rc7 #41 SMP PREEMPT Sat Sep 2 12:42:52 
CEST 2017 x86_64 GNU/Linux
Machine Type: i586-pc-linux-gnu

Bash Version: 4.3
Patch Level: 33
Release Status: release

Description:
As stated in the man page, bash will only allow a single coproc at a time.
Thus, when a first coproc has already ended, a second one can be created.
This can e.g. be verified by commenting out the upper #sleep line, below.
Alternatively, you can remove the parens that start the subshell.

But if we start a subshell while the coproc still runs, then wait inside 
the subshell
until the first coproc has ended and then try to start another coproc, 
bash prints the
infamous warning:
./test_sp1.sh: line 10: warning: execute_coproc: coproc [11030:COPROC] 
still exists

The process id 11030 does (verifyably - e.g. add `ps a`) not exist the 
time the warning is printed.░

Repeat-By:
coproc { echo "Start 1" >&2; sleep 1; echo "Stop 1" >&2; }

#sleep 1.1 # use this instead and the warning will disapped
sleep 0.9

(
sleep 0.5
coproc { echo "Start 2" >&2; sleep 1; echo "Stop 2" >&2; }
)

sleep 1.5

echo "Done"

Fix:
Probably some internal variables have to be reset on fork().