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

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:

quoted

On 2021-09-22 01:45:18 [+0200], Frederic Weisbecker wrote:

quoted

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