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Date: Fri, 6 Nov 2009 12:18:14 -0800
From: Matt Helsley <matthltc@us.ibm.com>
To: "Serge E. Hallyn" <serue@us.ibm.com>
Cc: Sukadev Bhattiprolu <sukadev@us.ibm.com>, arnd@arndb.de,
       Containers <containers@lists.linux-foundation.org>,
       linux-kernel@vger.kernel.org,
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       Alexey Dobriyan <adobriyan@gmail.com>, roland@redhat.com,
       Pavel Emelyanov <xemul@openvz.org>
Subject: Re: [v11][PATCH 9/9] Document clone_with_pids() syscall
Message-ID: <20091106201814.GA26614@count0.beaverton.ibm.com>
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 <20091106183936.GA32531@us.ibm.com>
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On Fri, Nov 06, 2009 at 12:39:36PM -0600, Serge E. Hallyn wrote:
> Quoting Sukadev Bhattiprolu (sukadev@us.ibm.com):
> ...
> > +	If a pid in the @pids list is non-zero, the kernel tries to assign
> > +	the specified pid in that namespace.  If that pid is already in use
> > +	by another process, the system call fails (see EBUSY below).
> > +
> > +	The order of pids in @pids is oldest in pids[0] to youngest pid
> > +	namespace in pids[nr_pids-1]. If the number of pids specified in the
> 
> In the sys_choosepid() discussion, Matt suggested it would be more
> user-friendly to have the pid for the youngest pidns be pids[0].
> That way the user doesn't have to know their pidns depth.

As far as I could see, Suka's solution also does not require knowing
the pidns depth (aka level). He made it so that copy_from_user()
adjusts its destination using the discrepancy between the number of
pids passed and the number of levels.

If userspace passes an array with n pids and there are k namespace levels
then clone_with_pids() makes sure that the kernel sees a pid array like:

index	  0     ... k - (n + 1)        ...          k - 1
	+-----------------------+-------------------------+
pid_t	| 0 ..................0 | <copied from userspace> |
	+-----------------------+-------------------------+

So even though the order is different from choosepid() the calling
task still doesn't need to know its pidns level. Of course, just
like choosepid(), n <= k or userspace will get EINVAL.

Cheers,
	-Matt Helsley

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