Problems with ALT_OUTPUTDIR in debug build

[Volker Simonis]

Hi,

me again. I' found that the additional "-fastdebug" directories are
created unconditionally by the top-level makefile, in the setup-step:

setup:
        $(MKDIR) -p $(OUTPUTDIR)/j2sdk-image
        $(MKDIR) -p $(ABS_OUTPUTDIR)/j2sdk-image
        $(MKDIR) -p $(OUTPUTDIR)-fastdebug/j2sdk-image
        $(MKDIR) -p $(ABS_OUTPUTDIR)-fastdebug/j2sdk-image

Is this really necessary?

Morever I found that I can create a debug build that honours the ALT_OUTPUTDIR
setting and leaves me with only one bogus "-fastdebug" directory as follows:

make SKIP_DEBUG_BUILD=false SKIP_FASTDEBUG_BUILD=true DEBUG_NAME=debug
ALT_OUTPUTDIR=/build/my_debug_build

This will create the following two directories in /build:

my_debug_build
my_debug_build-fastdebug

and put all the stuff into "my_debug_build" as desired. After the
build I can remove "my_debug_build-fastdebug" and be happy.

Any comments if this is the right way to do a debug build?

Regards,
Volker


On 1/10/08, Volker Simonis <volker.simonis at gmail.com> wrote:
> Hi,
> there seems to be a problem with the handling of ALT_OUTPUTDIR for debug and
> fastdebug builds. The top-level makefile contains the following definitions:
>
> Makefile
> --------
>
> include $(JDK_MAKE_SHARED_DIR)/Defs-control.gmk
>
> COMMON_DEBUG_FLAGS= \
>         DEBUG_NAME=$(DEBUG_NAME) \
>         ALT_OUTPUTDIR=$(_OUTPUTDIR)-$(DEBUG_NAME) \
>         NO_DOCS=true
>
> generic_debug_build:
>         @$(ECHO) $@ build started: `$(DATE) '+%y-%m-%d %H:%M'`
>         $(MAKE) $(COMMON_DEBUG_FLAGS) setup build
>         @$(ECHO) $@ build finished: `$(DATE) '+%y-%m-%d %H:%M'`
>
>
> As you can see, it always sets the value of ALT_OUTPUTDIR to
> "$(_OUTPUTDIR)-$(DEBUG_NAME)" where _OUTPUTDIR is the default value of the
> output directory which is defined in "jdk/make/common/shared/Defs-control.gmk"
> as follows (i.e. "linux-i586" on x86/Linux):
>
> jdk/make/common/shared/Defs-control.gmk
> ---------------------------------------
> # Default output directory
> _OUTPUTDIR=$(CONTROL_TOPDIR)/build/$(PLATFORM)-$(ARCH)
>
> Shouldn't COMMON_DEBUG_FLAGS use $(OUTPUTDIR) instead of $(_OUTPUTDIR) or
> am I missing something.
>
>
> It is also quite confusing that the build always creates
> $(OUTPUTDIR)-fastdebug subdirectories, no matter if we do a fastdebug build or
> not. I think that only the directories should be created which are used later
> on for the build. If I do for example "make debug_build" without specifying
> "ALT_OUTPUT_DIR", make will create four output directories as follows:
>
> linux-i586
> linux-i586-debug
> linux-i586-debug-fastdebug
> linux-i586-fastdebug
>
> in the build directory, while the output only goes to "linux-i586-debug".
> What are the "-fastdebug"-directories (especially
> "linux-i586-debug-fastdebug") good for?
>
> If I do "make debug_build ALT_OUTPUTDIR=/build/xxx" the build creates:
>
> linux-i586-debug
> linux-i586-debug-fastdebug
> xxx
> xxx-fastdebug
>
> but the output goes to "linux-i586-debug" as in the previous case.
>
> If we fix the top-level Makefile as suggested above, we'll get:
>
> xxx
> xxx-debug
> xxx-debug-fastdebug
> xxx-fastdebug
>
> with all the output going to "xxx-debug". That seems much better, but
> still we have three unused directories!
>
> In my eyes, the cleanest solution would be if ALT_OUTPUTDIR would be honoured
> "as is", as the output directory for everything we built, without anything
> appended to it. So the developer should be free to choose whatever he wants as
> the output directory. And there should be no additional directories created.
>
> Regards,
> Volker
>