Re: $? behavior difference between grouped targets and a multi-target pattern rule

[Masahiro Yamada]

On Sat, Aug 10, 2024 at 8:01 AM Dmitry Goncharov
<dgoncharov@users.sf.net> wrote:
>
> On Fri, Aug 9, 2024 at 8:07 AM Masahiro Yamada <masahiroy@kernel.org> wrote:
> > The value of $? is different.
>
> What do you observe if you wait between touching foo.r and touching foo.p?
>
> $ rm foo.*; touch foo.p; sleep 2; touch foo.r
> $ make -f makefile1
> $? = foo.r FORCE
> touch foo.p ; touch foo.q
> $
> $ rm foo.*; touch foo.p; sleep 2; touch foo.r
> $ make -f makefile2
> $? = foo.r FORCE
> touch foo.p ; touch foo.q
> $
> $
> $ rm foo.*; touch foo.r; sleep 2; touch foo.p
> $ make -f makefile1
> $? = FORCE
> touch foo.p ; touch foo.q
> $ rm foo.*; touch foo.r; sleep 2; touch foo.p
> $ make -f makefile2
> $? = FORCE
> touch foo.p ; touch foo.q
>
>
> regards, Dmitry



Which GNU Make version did you test?


As I mentioned, commit fabb03eac412b5ea19f1a97be31dc8c6fa7fc047
changed (fixed) the behavior, which introduced the inconsistency
between grouped targets and a multi-target pattern rule.




The first example you gave was never a problem because
foo.p is outdated and foo.q is missing.



The problematic case is
foo.p is up-to-date, and foo.q is missing.





[before fabb03eac412b5ea19f1a97be31dc8c6fa7fc047]



$ rm foo.*; touch foo.r; sleep 2; touch foo.p
$ make-4.3  -f Makefile1
$? = FORCE
touch foo.p ; touch foo.q

$ rm foo.*; touch foo.r; sleep 2; touch foo.p
$ make-4.3  -f Makefile2
$? = FORCE
touch foo.p ; touch foo.q







[after fabb03eac412b5ea19f1a97be31dc8c6fa7fc047]

$ rm foo.*; touch foo.r; sleep 2; touch foo.p
$ make-4.4  -f Makefile1
$? = foo.r FORCE
touch foo.p ; touch foo.q

$ rm foo.*; touch foo.r; sleep 2; touch foo.p
$ make-4.4  -f Makefile2
$? = FORCE
touch foo.p ; touch foo.q




I just thought fabb03eac412b5ea19f1a97be31dc8c6fa7fc047
addressed the issue in a consistent way
between grouped targets and a pattern rule.


When any of (foo.p, foo.q) is missing or out of date,
even if make is primarily trying to build foo.p
(because all: depends on foo.p and foo.q in this order),
we end up with rebuilding both foo.p and foo.q
simultaneously in a single recipe.


foo.r is a prerequisite newer than foo.q
(but older than foo.p), so I believe foo.r should
be contained in $?






-- 
Best Regards
Masahiro Yamada