Re: rcu/tree: Protect rcu_rdp_is_offloaded() invocations on RT

[Sebastian Andrzej Siewior]

On 2021-09-22 13:10:12 [+0200], Frederic Weisbecker wrote:
> On Wed, Sep 22, 2021 at 08:32:08AM +0200, Sebastian Andrzej Siewior wrote:
> > On 2021-09-22 01:45:18 [+0200], Frederic Weisbecker wrote:
> > > 
> > > Also while at it, I'm asking again: traditionally softirqs could assume that
> > > manipulating a local state was safe against !irq_count() code fiddling with
> > > the same state on the same CPU.
> > > 
> > > Now with preemptible softirqs, that assumption can be broken anytime. RCU was
> > > fortunate enough to have a warning for that. But who knows how many issues like
> > > this are lurking?
> > 
> > If "local state" is modified then it is safe as long as it is modified
> > within a local_bh_disable() section. And we are in this section while
> > invoking a forced-threaded interrupt. The special part about RCU is
> > that it is used in_irq() as part of core-code.
> 
> But local_bh_disable() was deemed for protecting from interrupting softirqs,
> not the other way around (softirqs being preempted by other tasks). The latter
> semantic is new and nobody had that in mind until softirqs have been made
> preemptible.
> 
> For example:
> 
>                              CPU 0
>           -----------------------------------------------
>           SOFTIRQ                            RANDOM TASK
>           ------                             -----------
>           int *X = &per_cpu(CPUX, 0)         int *X = &per_cpu(CPUX, 0)
>           int A, B;                          WRITE_ONCE(*X, 0);
>                                              WRITE_ONCE(*X, 1);
>           A = READ_ONCE(*X);
>           B = READ_ONCE(*X);
> 
> 
> We used to have the guarantee that A == B. That's not true anymore. Now
> some new explicit local_bh_disable() should be carefully placed on RANDOM_TASK
> where it wasn't necessary before. RCU is not that special in this regard.

The part with rcutree.use_softirq=0 on RT does not make it any better,
right?
So you rely on some implicit behaviour which breaks with RT such as:

                              CPU 0
           -----------------------------------------------
           RANDOM TASK-A                      RANDOM TASK-B
           ------                             -----------
           int *X = &per_cpu(CPUX, 0)         int *X = &per_cpu(CPUX, 0)
           int A, B;                          
					      spin_lock(&D);
           spin_lock(&C);
	   				      WRITE_ONCE(*X, 0);
           A = READ_ONCE(*X);
                                              WRITE_ONCE(*X, 1);
           B = READ_ONCE(*X);

while spinlock C and D are just random locks not related to CPUX but it
just happens that they are held at that time. So for !RT you guarantee
that A == B while it is not the case on RT.

Sebastian