linux-kernel - Re: [PATCH 2/3] ftrace: use struct pid

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Date:	Thu, 4 Dec 2008 09:29:04 -0500 (EST)
From:	Steven Rostedt <rostedt@...dmis.org>
To:	Dave Hansen <dave@...ux.vnet.ibm.com>
cc:	"Eric W. Biederman" <ebiederm@...ssion.com>,
	Frederic Weisbecker <fweisbec@...il.com>,
	linux-kernel@...r.kernel.org, Steven Rostedt <srostedt@...hat.com>,
	containers@...ts.osdl.org, Ingo Molnar <mingo@...e.hu>,
	Sukadev Bhattiprolu <sukadev@...ux.vnet.ibm.com>,
	Andrew Morton <akpm@...ux-foundation.org>
Subject: Re: [PATCH 2/3] ftrace: use struct pid


On Thu, 4 Dec 2008, Dave Hansen wrote:

> On Thu, 2008-12-04 at 04:56 -0800, Dave Hansen wrote:
> > On Thu, 2008-12-04 at 04:42 -0800, Eric W. Biederman wrote:
> > > 
> > > > +static void clear_ftrace_pid_task(struct pid **pid)
> > > > +{
> > > > +     struct task_struct *p;
> > > > +
> > >         rcu_read_lock();
> > > 
> > > > +     do_each_pid_task(*pid, PIDTYPE_PID, p) {
> > > > +             clear_tsk_trace_trace(p);
> > > > +     } while_each_pid_task(*pid, PIDTYPE_PID, p);
> > >         rcu_read_unlock()
> > > 
> > > > +     put_pid(*pid);
> > > > +
> > > > +     *pid = NULL;
> > > > +}
> > 
> > Could we get away with sticking the rcu_read_{un}lock() inside those
> > macros?  Those are going to get used in pretty high level code and we're
> > allowed to nest rcu_read_lock().  No danger of deadlocks or lock
> > inversions.
> 
> Why don't any of the other users of do_each_pid_task() use
> rcu_read_lock()?  They all seem to be under read_lock(&tasklist_lock)
> (except one is under a write lock of the same).

Well, if the pid hashes are traversal safe (rcu style), then we only worry 
about a node or task being freed. I'm assuming that the node is protected 
via RCU as tasks are, then using only rcu_read_lock should be sufficient.

-- Steve

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