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

From: Mike Hommey <mh@glandium.org>
Date: Jan 22, 2015, 15:09 UTC
Message-ID: <20150122150914.GA20135@glandium.org>
In-Reply-To: <20150122130036.GC19681@peff.net>

On Thu, Jan 22, 2015 at 08:00:36AM -0500, Jeff King wrote:

Show 26 quoted lines

> 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?

Maybe this comes from the misconception that debugging information changes the generated code, which, in fact, it doesn't.

  $ make CFLAGS=-g LDFLAGS=-Wl,--build-id=none >/dev/null 2>&1 && wc -c <git
  4432768
  $ strip --strip-debug git && wc -c < git
  2391120
  $ cp git git_
  $ make -j4 CFLAGS= LDFLAGS=-Wl,--build-id=none >/dev/null 2>&1 && wc -c <git
  2400192
  $ strip --strip-debug git && wc -c < git
  2391120
  $ diff -s git git_
  Files git and git_ are identical

LDFLAGS=-Wl,--build-id=none just avoids creating a .note.gnu.build-id section containing a uuid that varies between builds. The 9k difference between unstripped vs stripped for the no-debug-info case comes from the removal of the few symbols for source file names (all the symbols from readelf -s git | grep ABS).

Mike