Re: [RFC][PATCH] Make access to taks's nsproxy liter

[Oleg Nesterov <oleg-6lXkIZvqkOAvJsYlp49lxw@public.gmane.org>]

On 08/09, Pavel Emelyanov wrote:
>
> Paul E. McKenney wrote:
> >On Wed, Aug 08, 2007 at 08:41:07PM +0400, Oleg Nesterov wrote:
> >>
> >>>+void switch_task_namespaces(struct task_struct *p, struct nsproxy *new)
> >>>+{
> >>>+	struct nsproxy *ns;
> >>>+
> >>>+	might_sleep();
> >>>+
> >>>+	ns = p->nsproxy;
> >>>+	if (ns == new)
> >>>+		return;
> >>>+
> >>>+	if (new)
> >>>+		get_nsproxy(new);
> >>>+	rcu_assign_pointer(p->nsproxy, new);
> >>>+
> >>>+	if (ns && atomic_dec_and_test(&ns->count)) {
> >>>+		/*
> >>>+		 * wait for others to get what they want from this
> >>>+		 * nsproxy. cannot release this nsproxy via the
> >>>+		 * call_rcu() since put_mnt_ns will want to sleep
> >>>+		 */
> >>>+		synchronize_rcu();
> >>>+		free_nsproxy(ns);
> >>>+	}
> >>>+}
> >>(I may be wrong, Paul cc'ed)
> >>
> >>This is correct with the current implementation of RCU, but strictly 
> >>speaking,
> >>we can't use synchronize_rcu() here, because write_lock_irq() doesn't 
> >>imply
> >>rcu_read_lock() in theory.
> >
> >Can you use synchronize_sched() instead?  The synchronize_sched()
> 
> #define synchronize_sched() synchronize_rcu()
> they are the same? what's the point?

There are the same with the current implementation. RT kernel for example,
has another, when preempt_disable() doesn't imply rcu_read_lock().

> >primitive will wait until all preempt/irq-disable code sequences complete.
> >Therefore, it would wait for all write_lock_irq() code sequences to
> >complete.
> 
> But we don't need this. Iff we get the nsproxy under rcu_read_lock() all
> we need is to wait for RCU sections to complete.

Yes. But this patch complicates the code and slows down group_exit. We don't
access non-current ->nsproxy so often afaics, and task_lock is cheap.

Note also that switch_task_namespaces() might_sleep(), but sys_unshare()
calls it under task_lock().

Oleg.