Re: [PATCH] Makefile: do not compile git with debugging symbols by default

> 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@peff.net>:
>> 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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