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Date:      Tue, 25 Oct 2011 09:40:34 +0100
From:      "Robert N. M. Watson" <rwatson@freebsd.org>
To:        Kostik Belousov <kostikbel@gmail.com>
Cc:        Mikolaj Golub <trociny@freebsd.org>, freebsd-hackers@freebsd.org
Subject:   Re: "ps -e" without procfs(5)
Message-ID:  <7292FA7A-60A4-4D29-87B5-894AF0ED9C0E@freebsd.org>
In-Reply-To: <20111025082451.GO50300@deviant.kiev.zoral.com.ua>
References:  <86y5wkeuw9.fsf@kopusha.home.net> <20111016171005.GB50300@deviant.kiev.zoral.com.ua> <86aa8qozyx.fsf@kopusha.home.net> <20111025082451.GO50300@deviant.kiev.zoral.com.ua>

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On 25 Oct 2011, at 09:24, Kostik Belousov wrote:

> On Tue, Oct 25, 2011 at 12:13:10AM +0300, Mikolaj Golub wrote:
>>=20
>> On Sun, 16 Oct 2011 20:10:05 +0300 Kostik Belousov wrote:
>>=20
>> KB> In my opinion, the way to implement the feature is to (re)use
>> KB> linprocfs_doargv() and provide another kern.proc sysctl to =
retrieve the
>> KB> argv and env vectors. Then, ps(1) and procstat(1) can use it, as =
well as
>> KB> procfs and linprocfs inside the kernel.
>>=20
>> Thanks! I am testing a patch (without auxv vector so far) and have =
some
>> questions.
>>=20
>> Original ps -e returns environment only for user owned processes (the =
access is
>> restricted by the permissions of /proc/pid/mem file). My =
kern.proc.env sysctl
>> does not have such a restriction. I suppose I should add it? What =
function I
>> could use for this?
>>=20
>> BTW, linprocfs allows to read other user's environment.
> linprocfs uses p_cansee() to check the permissions. There are sysctls
> security.bsd.see_other_{ug}ids that control the behaviour.
>=20
> I believe that the new sysctl shall use the same check.

To be honest, I'd be far more comfortable if the environment check used =
p_candebug(). Environmental variables sometimes contain passwords, etc, =
that shouldn't be visible to other users on the system. Even showing =
command lines is a bit dubious, but widely accepted, whereas seeing the =
contents of environmental variables is not widely known in user =
communities.

Robert=



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