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
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