Hello Jeff,
Yes, main point is size of executable. I'm sorry, didn't see 'strip' target.
What if we will strip git and other executable files by default and
add -g option and don't strip it if DEBUG=1 passed to make. Something
like this:
git$X: git.o GIT-LDFLAGS $(BUILTIN_OBJS) $(GITLIBS)
$(QUIET_LINK)$(CC) $(ALL_CFLAGS) -o $@ $(ALL_LDFLAGS) git.o \
$(BUILTIN_OBJS) $(LIBS)
ifneq ($(DEBUG),1)
$(MAKE) strip
endif
Thank you.
2015-01-22 19:00 GMT+06:00 Jeff King <peff@xxxxxxxx>:
> On Thu, Jan 22, 2015 at 06:50:37PM +0600, Alexander Kuleshov wrote:
>
>> Standard user has no need in debugging information. This patch adds
>> DEBUG=1 option to compile git with debugging symbols and compile without
>> it by default.
>
> This explanation is missing why it is beneficial _not_ to have the
> debugging information.
>
> I expect the answer is "it makes the executable smaller". And that is
> true, but it gets smaller still if you run "strip" on the result:
>
> $ make CFLAGS= >/dev/null 2>&1 && wc -c <git
> 2424248
>
> $ make CFLAGS=-g >/dev/null 2>&1 && wc -c <git
> 4500816
>
> $ strip git && wc -c <git
> 2109200
>
> So I am not sure who this is helping. If you are size-conscious, you
> should use strip, in which case the "-g" flag does not matter (and we
> even have "make strip" to help you).
>
> Is there some other reason to avoid the debugging information?
>
> -Peff
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