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

[Hongyi Zhao]

On Wed, Jan 19, 2022 at 2:56 AM Paul Smith <psmith@gnu.org> wrote:
>
> On Tue, 2022-01-18 at 10:14 +0800, Hongyi Zhao wrote:
> > But it seems that the `$?' used above can't obtain the exit status of
> > the code block in @(...) when used from outside the @() structure
> > [2].  So, I want to know if I can reliably obtain the exit status of
> > the code block in @(...) from outside the @() structure. Any hints
> > will be greatly appreciated.
>
> First, of course you must use $$? not $?, because $? is a shell
> variable not a make variable so you must escape it from make.
>
> Second, there's nothing magical about ().

So, I would like to know when I should use () to group some commands
and when not.

> All versions of make, including GNU make, invoke each logical line of
> the recipe in a separate shell.  There is no interaction between two
> different shells: the exit code of one shell cannot be accessed by a
> sibling shell.
>
> A rule like this:
>
>     foo:
>             (exit 1)
>             echo $$?
>
> causes make to invoke two different shells:
>
>   /bin/sh -c '(exit 1)'
>   /bin/sh -c 'echo $?'
>
> The exit code of the first shell is not put into the $? variable of the
> second shell.
>
> If you want to access the exit code of a previous command you must put
> them both into the same shell, which means they both need to be in the
> same logical line of the recipe.  So, this rule:
>
>     foo:
>             (exit 1) ; \
>             echo $$?
>
> invokes a single shell command:
>
>   /bin/sh -c '(exit 1) ; echo $?'
>
> now it will work as you expect.

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?


> So, in your makefile rule, you just need to make sure that you add
> semicolons and backslashes to ensure that the recipe is treated as one
> logical line; change:
>
>   echo "*** located here $(2)" ; \
>   exit 1 ; fi ; fi ; fi)
>   if test $? -eq 0 -a ! -e ../$(3); then \
>
> to add the semicolon / backslash:
>
>   echo "*** located here $(2)" ; \
>   exit 1 ; fi ; fi ; fi) ; \
>   if test $$? -eq 0 -a ! -e ../$(3); then \

Thank you for your correction.

HZ