IF YOU WOULD LIKE TO GET AN ACCOUNT, please write an
email to Administrator. User accounts are meant only to access repo
and report issues and/or generate pull requests.
This is a purpose-specific Git hosting for
BaseALT
projects. Thank you for your understanding!
Только зарегистрированные пользователи имеют доступ к сервису!
Для получения аккаунта, обратитесь к администратору.
Bash 4.4, released in 2016, supports 'wait $!' to check the exit status
of a process substitution, but it seems too new.
Some people using older bash versions (on CentOS 7, Ubuntu 16.04, etc.)
reported an error like this:
./scripts/check-local-export: line 54: wait: pid 17328 is not a child of this shell
I used the process substitution to avoid a pipeline, which executes each
command in a subshell. If the while-loop is executed in the subshell
context, variable changes within are lost after the subshell terminates.
Fortunately, Bash 4.2, released in 2011, supports the 'lastpipe' option,
which makes the last element of a pipeline run in the current shell process.
Switch to the pipeline with 'lastpipe' solution, and also set 'pipefail'
to catch errors from ${NM}.
Add the bash requirement to Documentation/process/changes.rst.
Fixes: 31cb50b559 ("kbuild: check static EXPORT_SYMBOL* by script instead of modpost")
Reported-by: Tetsuo Handa <penguin-kernel@I-love.SAKURA.ne.jp>
Reported-by: Michael Ellerman <mpe@ellerman.id.au>
Reported-by: Wang Yugui <wangyugui@e16-tech.com>
Tested-by: Tetsuo Handa <penguin-kernel@I-love.SAKURA.ne.jp>
Tested-by: Jon Hunter <jonathanh@nvidia.com>
Acked-by: Nick Desaulniers <ndesaulniers@google.com>
Tested-by: Sedat Dilek <sedat.dilek@gmail.com> # LLVM-14 (x86-64)
Signed-off-by: Masahiro Yamada <masahiroy@kernel.org>
da4288b95b