Re: lockdep WARNING on check_critical_timing()

["Paul E. McKenney" <paulmck@linux.vnet.ibm.com>]

On Thu, Sep 06, 2012 at 05:56:39PM -0400, Steven Rostedt wrote:
> On Thu, 2012-09-06 at 14:29 -0700, Paul E. McKenney wrote:
> >  
> > > Didn't we talk about having the rcu_dereference_raw() not do the check?
> > > The function tracer is just too invasive to add work arounds to prevent
> > > lockdep from screaming about it.
> > 
> > Actually, rcu_dereference_raw() is already supposed to bypass the
> > lockdep checks.  And the code looks to me like it does the bypass,
> > OR-ing "1" into the asssertion condition.
> > 
> > So what am I missing here?
> 
> >From my tree, I see:
> 
> #define rcu_dereference_raw(p) rcu_dereference_check(p, 1)
> 
> #define rcu_dereference_check(p, c) \
> 	__rcu_dereference_check((p), rcu_read_lock_held() || (c), __rcu)
> 
> Note the 'c' comes after rcu_read_lock_held()
> 
> static inline int rcu_read_lock_held(void)
> {
> 	if (!debug_lockdep_rcu_enabled())
> 		return 1;
> 	if (rcu_is_cpu_idle())
> 		return 0;
> 	if (!rcu_lockdep_current_cpu_online())
> 		return 0;
> 	return lock_is_held(&rcu_lock_map);
> }
> 
> Then when lock_is_held() is called, we get the false warning message.

OK, I can easily do:

	__rcu_dereference_check((p), (c) || rcu_read_lock_held(), __rcu)

But I am still missing why the order matters.  Are you saying that
lock_is_held() itself is doing the splat?

							Thanx, Paul