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

From: David Aguilar <davvid@gmail.com>
Date: Jan 27, 2015, 08:43 UTC
Message-ID: <20150127084338.GA37443@gmail.com>
In-Reply-To: <20150122225517.GB31912@glandium.org>

On Fri, Jan 23, 2015 at 07:55:17AM +0900, Mike Hommey wrote:

Show 15 quoted lines

> On Thu, Jan 22, 2015 at 01:35:38PM -0500, Jeff King wrote:
> > On Thu, Jan 22, 2015 at 06:36:41PM +0100, Matthieu Moy wrote:
> > 
> > > > Yes, main point is size of executable.
> > > 
> > > The Git executable is a few megabytes, i.e. 0.001% the size of a really
> > > small hard disk. The benefit seems really negligible to me.
> > 
> > I don't know the layout of the symbols with respect to the code, or
> > whether the stripped version might reduce memory pressure. So in theory
> > it could have a performance impact.
> 
> It doesn't. Debugging info is in a part of the file that is not mapped
> in memory, and in a part that can be removed without affecting the rest
> of the file, so it's more or less at the end.

It goes even further. These days Fedora systems strip debug info out into separate files and packages while creating rpms debuginfo packages are created automatically and provide debuginfo files under /usr/lib/debug, where gdb knows to look by default.

Alexander, one nice thing about the Makefile is that it supports you creating a file in your Git worktree called "config.mak" with the following content:

CFLAGS = -O2 -Wall

If you do that then git will build without debug info and you won't have to specify CFLAGS when invoking "make". Hopefully that's easy and convenient enough.

cheers,

-- 
David