linux-kernel - Re: LKMM litmus test for Roman Penyaev's rcu-rr

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Date:   Wed, 6 Jun 2018 16:41:20 +0200
From:   Roman Penyaev <roman.penyaev@...fitbricks.com>
To:     Alan Stern <stern@...land.harvard.edu>
Cc:     "Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>,
        Linus Torvalds <torvalds@...ux-foundation.org>,
        Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
        linux-arch <linux-arch@...r.kernel.org>,
        andrea.parri@...rulasolutions.com,
        Will Deacon <will.deacon@....com>,
        Peter Zijlstra <peterz@...radead.org>,
        Boqun Feng <boqun.feng@...il.com>,
        Nick Piggin <npiggin@...il.com>,
        David Howells <dhowells@...hat.com>,
        Jade Alglave <j.alglave@....ac.uk>,
        Luc Maranget <luc.maranget@...ia.fr>,
        Akira Yokosawa <akiyks@...il.com>,
        Ingo Molnar <mingo@...nel.org>
Subject: Re: LKMM litmus test for Roman Penyaev's rcu-rr

On Wed, Jun 6, 2018 at 3:54 PM, Alan Stern <stern@...land.harvard.edu> wrote:
> On Wed, 6 Jun 2018, Roman Penyaev wrote:
>
>> > Preserving the order of volatile accesses isn't sufficient.  The
>> > compiler is still allowed to translate
>> >
>> >         r1 = READ_ONCE(x);
>> >         if (r1) {
>> >                 ...
>> >         }
>> >         WRITE_ONCE(y, r2);
>> >
>> > into something resembling
>> >
>> >         r1 = READ_ONCE(x);
>> >         WRITE_ONCE(y, r2);
>> >         if (r1) {
>> >                 ...
>> >         }
>>
>> Hi Alan,
>>
>> According to the standard C99 Annex C "the controlling expression of
>> a selection statement (if or switch)" are the sequence points, just
>> like a volatile access (READ_ONCE or WRITE_ONCE).
>>
>> "5.1.2.3 Program execution" states:
>> "At certain specified points in the execution sequence called sequence
>> points, all side effects of previous evaluations shall be complete
>> and no side effects of subsequent evaluations shall have taken place."
>>
>> So in the example we have 3 sequence points: "READ_ONCE", "if" and
>> "WRITE_ONCE", which it seems can't be reordered.  Am I mistaken
>> interpreting standard?  Could you please clarify.
>
> Well, for one thing, we're talking about C11, not C99.

C11 is a n1570, ISO/IEC 9899:2011 ? (according to wiki).  Found pdf on
the web contains similar lines, so should not be any differences for
that particular case.

> For another, as far as I understand it, the standard means the program
> should behave _as if_ the side effects are completed in the order
> stated.  It doesn't mean that the generated code has to behave that way
> literally.

Then I do not understand what are the differences between "side effects
are completed" and "code generated".  Abstract machine state should
provide some guarantees between sequence points, e.g.:

    foo();    /* function call */
    ------------|
    *a = 1;     |
    *b = 12;    | Compiler in his right to reorder.
    *c = 123;   |
    ------------|
    boo();    /* function call */

compiler in his right to reorder memory accesses between foo() and
boo() calls (foo and boo are sequence points, but memory accesses
are not), but:

   foo();    /* function call */
   *(volatile int *)a = 1;
   *(volatile int *)b = 12;
   *(volatile int *)c = 123;
   boo();    /* function call */

are all sequence points, so compiler can't reorder them.

Where am I mistaken?

> And in particular, the standard is referring to the
> behavior of a single thread, not the interaction between multiple
> concurrent threads.

Yes, that is clear: we are talking about code reordering in one
particular function in a single threaded environment.

--
Roman