Re: Can the exit status of the code block in @(...) be obtained from outside the @() structure?

[Paul Smith]

On Wed, 2022-01-19 at 09:50 +0800, Hongyi Zhao wrote:
> So, I would like to know when I should use () to group some commands
> and when not.

Your question is about how to write shell scripts, not about how to
write makefiles.

The recipe of a makefile is a shell script.  You should use (), which
(in this situation) is a shell script construct, when you would want to
use a subshell (which is what this syntax provides) in a shell script. 
It has nothing at all to do with make.

The help-make mailing list is probably not the best place to learn how
to write shell scripts or understand them.

> But see the following testings:
> 
> werner@X10DAi-00:~$ /bin/sh -c '(exit 1); echo $?'
> 1
> werner@X10DAi-00:~$ /bin/sh -c 'exit 1; echo $?'
> werner@X10DAi-00:~$
> 
> Why must I use () here, otherwise, the exit code will not be
> captured?

Again, this is a shell script question (you can tell, because you are
not running make at all here) and your best bet is to find a group or
manual or tutorial that can help you understand shell scripting.

It's not correct to say it "will not be captured".  It is captured. For
example you can run:

  $ /bin/sh -c 'exit 1; echo $?'
  $ echo $?
  1

What's correct is that it is not PRINTED.

The "exit" command in a shell script exits from the shell script.  So
when you run:

   /bin/sh -c 'exit 1; echo $?'

the exit command causes the shell script to exit when it runs the exit
command.  So, it never reaches the echo command because it exited
first.

The () adds a subshell, so when you run:

  /bin/sh -c '(exit 1); echo $?'

that's similar to running this:

  /bin/sh -c '/bin/sh -c "exit 1"; echo $?'

Now the exit command exits from the inner /bin/sh but not the outer
/bin/sh, so then the outer /bin/sh runs the echo command.