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Message-ID: <20201110152124.GR3249@paulmck-ThinkPad-P72>
Date: Tue, 10 Nov 2020 07:21:24 -0800
From: "Paul E. McKenney" <paulmck@...nel.org>
To: Stephen Rothwell <sfr@...b.auug.org.au>
Cc: Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
Linux Next Mailing List <linux-next@...r.kernel.org>,
Peter Zijlstra <peterz@...radead.org>,
Mauro Carvalho Chehab <mchehab+huawei@...nel.org>,
Jonathan Corbet <corbet@....net>
Subject: Re: linux-next: build warnings after merge of the rcu tree
On Tue, Nov 10, 2020 at 05:59:52PM +1100, Stephen Rothwell wrote:
> Hi all,
>
> After merging the rcu tree, today's linux-next build (htmldocs)
> produced these warnings:
>
> Documentation/RCU/Design/Requirements/Requirements.rst:119: WARNING: Malformed table.
My bad, apologies, queuing an alleged fix.
Thanx, Paul
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | Wait a minute! You said that updaters can make useful forward |
> | progress concurrently with readers, but pre-existing readers will |
> | block synchronize_rcu()!!! |
> | Just who are you trying to fool??? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | First, if updaters do not wish to be blocked by readers, they can use |
> | call_rcu() or kfree_rcu(), which will be discussed later. |
> | Second, even when using synchronize_rcu(), the other update-side |
> | code does run concurrently with readers, whether pre-existing or not. |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:178: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | Why is the synchronize_rcu() on line 28 needed? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | Without that extra grace period, memory reordering could result in |
> | do_something_dlm() executing do_something() concurrently with |
> | the last bits of recovery(). |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:289: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | But rcu_assign_pointer() does nothing to prevent the two |
> | assignments to ``p->a`` and ``p->b`` from being reordered. Can't that |
> | also cause problems? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | No, it cannot. The readers cannot see either of these two fields |
> | until the assignment to ``gp``, by which time both fields are fully |
> | initialized. So reordering the assignments to ``p->a`` and ``p->b`` |
> | cannot possibly cause any problems. |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:430: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | Without the rcu_dereference() or the rcu_access_pointer(), |
> | what destructive optimizations might the compiler make use of? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | Let's start with what happens to do_something_gp() if it fails to |
> | use rcu_dereference(). It could reuse a value formerly fetched |
> | from this same pointer. It could also fetch the pointer from ``gp`` |
> | in a byte-at-a-time manner, resulting in *load tearing*, in turn |
> | resulting a bytewise mash-up of two distinct pointer values. It might |
> | even use value-speculation optimizations, where it makes a wrong |
> | guess, but by the time it gets around to checking the value, an |
> | update has changed the pointer to match the wrong guess. Too bad |
> | about any dereferences that returned pre-initialization garbage in |
> | the meantime! |
> | For remove_gp_synchronous(), as long as all modifications to |
> | ``gp`` are carried out while holding ``gp_lock``, the above |
> | optimizations are harmless. However, ``sparse`` will complain if you |
> | define ``gp`` with ``__rcu`` and then access it without using either |
> | rcu_access_pointer() or rcu_dereference(). |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:513: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | Given that multiple CPUs can start RCU read-side critical sections at |
> | any time without any ordering whatsoever, how can RCU possibly tell |
> | whether or not a given RCU read-side critical section starts before a |
> | given instance of synchronize_rcu()? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | If RCU cannot tell whether or not a given RCU read-side critical |
> | section starts before a given instance of synchronize_rcu(), then |
> | it must assume that the RCU read-side critical section started first. |
> | In other words, a given instance of synchronize_rcu() can avoid |
> | waiting on a given RCU read-side critical section only if it can |
> | prove that synchronize_rcu() started first. |
> | A related question is “When rcu_read_lock() doesn't generate any |
> | code, why does it matter how it relates to a grace period?” The |
> | answer is that it is not the relationship of rcu_read_lock() |
> | itself that is important, but rather the relationship of the code |
> | within the enclosed RCU read-side critical section to the code |
> | preceding and following the grace period. If we take this viewpoint, |
> | then a given RCU read-side critical section begins before a given |
> | grace period when some access preceding the grace period observes the |
> | effect of some access within the critical section, in which case none |
> | of the accesses within the critical section may observe the effects |
> | of any access following the grace period. |
> | |
> | As of late 2016, mathematical models of RCU take this viewpoint, for |
> | example, see slides 62 and 63 of the `2016 LinuxCon |
> | EU <http://www2.rdrop.com/users/paulmck/scalability/paper/LinuxMM.201 |
> | 6.10.04c.LCE.pdf>`__ |
> | presentation. |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:548: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | The first and second guarantees require unbelievably strict ordering! |
> | Are all these memory barriers *really* required? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | Yes, they really are required. To see why the first guarantee is |
> | required, consider the following sequence of events: |
> | |
> | #. CPU 1: rcu_read_lock() |
> | #. CPU 1: ``q = rcu_dereference(gp); /* Very likely to return p. */`` |
> | #. CPU 0: ``list_del_rcu(p);`` |
> | #. CPU 0: synchronize_rcu() starts. |
> | #. CPU 1: ``do_something_with(q->a);`` |
> | ``/* No smp_mb(), so might happen after kfree(). */`` |
> | #. CPU 1: rcu_read_unlock() |
> | #. CPU 0: synchronize_rcu() returns. |
> | #. CPU 0: ``kfree(p);`` |
> | |
> | Therefore, there absolutely must be a full memory barrier between the |
> | end of the RCU read-side critical section and the end of the grace |
> | period. |
> | |
> | The sequence of events demonstrating the necessity of the second rule |
> | is roughly similar: |
> | |
> | #. CPU 0: ``list_del_rcu(p);`` |
> | #. CPU 0: synchronize_rcu() starts. |
> | #. CPU 1: rcu_read_lock() |
> | #. CPU 1: ``q = rcu_dereference(gp);`` |
> | ``/* Might return p if no memory barrier. */`` |
> | #. CPU 0: synchronize_rcu() returns. |
> | #. CPU 0: ``kfree(p);`` |
> | #. CPU 1: ``do_something_with(q->a); /* Boom!!! */`` |
> | #. CPU 1: rcu_read_unlock() |
> | |
> | And similarly, without a memory barrier between the beginning of the |
> | grace period and the beginning of the RCU read-side critical section, |
> | CPU 1 might end up accessing the freelist. |
> | |
> | The “as if” rule of course applies, so that any implementation that |
> | acts as if the appropriate memory barriers were in place is a correct |
> | implementation. That said, it is much easier to fool yourself into |
> | believing that you have adhered to the as-if rule than it is to |
> | actually adhere to it! |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:597: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | You claim that rcu_read_lock() and rcu_read_unlock() generate |
> | absolutely no code in some kernel builds. This means that the |
> | compiler might arbitrarily rearrange consecutive RCU read-side |
> | critical sections. Given such rearrangement, if a given RCU read-side |
> | critical section is done, how can you be sure that all prior RCU |
> | read-side critical sections are done? Won't the compiler |
> | rearrangements make that impossible to determine? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | In cases where rcu_read_lock() and rcu_read_unlock() generate |
> | absolutely no code, RCU infers quiescent states only at special |
> | locations, for example, within the scheduler. Because calls to |
> | schedule() had better prevent calling-code accesses to shared |
> | variables from being rearranged across the call to schedule(), if |
> | RCU detects the end of a given RCU read-side critical section, it |
> | will necessarily detect the end of all prior RCU read-side critical |
> | sections, no matter how aggressively the compiler scrambles the code. |
> | Again, this all assumes that the compiler cannot scramble code across |
> | calls to the scheduler, out of interrupt handlers, into the idle |
> | loop, into user-mode code, and so on. But if your kernel build allows |
> | that sort of scrambling, you have broken far more than just RCU! |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:738: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | Can't the compiler also reorder this code? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | No, the volatile casts in READ_ONCE() and WRITE_ONCE() |
> | prevent the compiler from reordering in this particular case. |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:793: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | Suppose that synchronize_rcu() did wait until *all* readers had |
> | completed instead of waiting only on pre-existing readers. For how |
> | long would the updater be able to rely on there being no readers? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | For no time at all. Even if synchronize_rcu() were to wait until |
> | all readers had completed, a new reader might start immediately after |
> | synchronize_rcu() completed. Therefore, the code following |
> | synchronize_rcu() can *never* rely on there being no readers. |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:1087: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | What about sleeping locks? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | These are forbidden within Linux-kernel RCU read-side critical |
> | sections because it is not legal to place a quiescent state (in this |
> | case, voluntary context switch) within an RCU read-side critical |
> | section. However, sleeping locks may be used within userspace RCU |
> | read-side critical sections, and also within Linux-kernel sleepable |
> | RCU `(SRCU) <#Sleepable%20RCU>`__ read-side critical sections. In |
> | addition, the -rt patchset turns spinlocks into a sleeping locks so |
> | that the corresponding critical sections can be preempted, which also |
> | means that these sleeplockified spinlocks (but not other sleeping |
> | locks!) may be acquire within -rt-Linux-kernel RCU read-side critical |
> | sections. |
> | Note that it *is* legal for a normal RCU read-side critical section |
> | to conditionally acquire a sleeping locks (as in |
> | mutex_trylock()), but only as long as it does not loop |
> | indefinitely attempting to conditionally acquire that sleeping locks. |
> | The key point is that things like mutex_trylock() either return |
> | with the mutex held, or return an error indication if the mutex was |
> | not immediately available. Either way, mutex_trylock() returns |
> | immediately without sleeping. |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:1295: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | Why does line 19 use rcu_access_pointer()? After all, |
> | call_rcu() on line 25 stores into the structure, which would |
> | interact badly with concurrent insertions. Doesn't this mean that |
> | rcu_dereference() is required? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | Presumably the ``->gp_lock`` acquired on line 18 excludes any |
> | changes, including any insertions that rcu_dereference() would |
> | protect against. Therefore, any insertions will be delayed until |
> | after ``->gp_lock`` is released on line 25, which in turn means that |
> | rcu_access_pointer() suffices. |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:1351: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | Earlier it was claimed that call_rcu() and kfree_rcu() |
> | allowed updaters to avoid being blocked by readers. But how can that |
> | be correct, given that the invocation of the callback and the freeing |
> | of the memory (respectively) must still wait for a grace period to |
> | elapse? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | We could define things this way, but keep in mind that this sort of |
> | definition would say that updates in garbage-collected languages |
> | cannot complete until the next time the garbage collector runs, which |
> | does not seem at all reasonable. The key point is that in most cases, |
> | an updater using either call_rcu() or kfree_rcu() can proceed |
> | to the next update as soon as it has invoked call_rcu() or |
> | kfree_rcu(), without having to wait for a subsequent grace |
> | period. |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:1893: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | Wait a minute! Each RCU callbacks must wait for a grace period to |
> | complete, and rcu_barrier() must wait for each pre-existing |
> | callback to be invoked. Doesn't rcu_barrier() therefore need to |
> | wait for a full grace period if there is even one callback posted |
> | anywhere in the system? |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | Absolutely not!!! |
> | Yes, each RCU callbacks must wait for a grace period to complete, but |
> | it might well be partly (or even completely) finished waiting by the |
> | time rcu_barrier() is invoked. In that case, rcu_barrier() |
> | need only wait for the remaining portion of the grace period to |
> | elapse. So even if there are quite a few callbacks posted, |
> | rcu_barrier() might well return quite quickly. |
> | |
> | So if you need to wait for a grace period as well as for all |
> | pre-existing callbacks, you will need to invoke both |
> | synchronize_rcu() and rcu_barrier(). If latency is a concern, |
> | you can always use workqueues to invoke them concurrently. |
> +-----------------------------------------------------------------------+
> Documentation/RCU/Design/Requirements/Requirements.rst:2220: WARNING: Malformed table.
>
> +-----------------------------------------------------------------------+
> | **Quick Quiz**: |
> +-----------------------------------------------------------------------+
> | But what if my driver has a hardware interrupt handler that can run |
> | for many seconds? I cannot invoke schedule() from an hardware |
> | interrupt handler, after all! |
> +-----------------------------------------------------------------------+
> | **Answer**: |
> +-----------------------------------------------------------------------+
> | One approach is to do ``rcu_irq_exit();rcu_irq_enter();`` every so |
> | often. But given that long-running interrupt handlers can cause other |
> | problems, not least for response time, shouldn't you work to keep |
> | your interrupt handler's runtime within reasonable bounds? |
> +-----------------------------------------------------------------------+
>
> Introduced by commit
>
> c0a41bf9dbc7 ("docs: Remove redundant "``" from Requirements.rst")
>
> --
> Cheers,
> Stephen Rothwell
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