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Date:      Thu, 27 Apr 2006 10:58:53 -0600
From:      Scott Long <scottl@samsco.org>
To:        Brooks Davis <brooks@one-eyed-alien.net>
Cc:        John-Mark Gurney <jmg@FreeBSD.org>, cvs-src@FreeBSD.org, src-committers@FreeBSD.org, Pawel Jakub Dawidek <pjd@FreeBSD.org>, cvs-all@FreeBSD.org
Subject:   Re: cvs commit: src/sys/kern sched_4bsd.c
Message-ID:  <4450F84D.4080601@samsco.org>
In-Reply-To: <20060427163351.GB12971@odin.ac.hmc.edu>
References:  <200604261942.k3QJgdnb086990@repoman.freebsd.org> <20060427102026.GC20828@garage.freebsd.pl> <20060427163351.GB12971@odin.ac.hmc.edu>

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Brooks Davis wrote:
> On Thu, Apr 27, 2006 at 12:20:26PM +0200, Pawel Jakub Dawidek wrote:
> 
>>On Wed, Apr 26, 2006 at 07:42:39PM +0000, John-Mark Gurney wrote:
>>+> jmg         2006-04-26 19:42:38 UTC
>>+> 
>>+>   FreeBSD src repository
>>+> 
>>+>   Modified files:
>>+>     sys/kern             sched_4bsd.c 
>>+>   Log:
>>+>   move remaining sysctl into the kern.sched tree...
>>[...]
>>+> -SYSCTL_INT(_kern, OID_AUTO, ccpu, CTLFLAG_RD, &ccpu, 0, "");
>>+> +SYSCTL_INT(_kern_sched, OID_AUTO, ccpu, CTLFLAG_RD, &ccpu, 0, "");
>>[...]
>>
>>Too fast.
>>
>># grep -r kern.ccpu /usr/src/
> 
> 
> We should retain the old location as well under COMPAT_FREEBSD[456].  It
> doesn't cost anything and we should treat the sysctl tree like any other
> API to the extent possible.
> 
> -- Brooks
> 
> 

Well, sysctl's are a second class API.  The ease by they can be changed, 
as well as the completely uncontrolled nature that they are developed 
in, means that we really start to hamstring ourselves if we declare them
as a first class API.  What we should do instead is define a subset of
the sysctl tree that is stable and guaranteed to exist, and declare that
the rest are transient and therefore not part of the API.  Doing this by
popular vote is impractical, what we need is someone to sit down and 
audit what is there now and make appropriate recommendations.  sysctls 
that are used by libkvm are a good candidate, for example, but random
driver sysctls might not be.

Scott




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