Re: [msysGit] Re: [PATCH/RFC] mingw: implement PTHREAD_MUTEX_INITIALIZER

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On Thu, Oct 27, 2011 at 19:00, Atsushi Nakagawa <atnak@xxxxxxxxx> wrote:
> Erik Faye-Lund <kusmabite@xxxxxxxxx> wrote:
>> On Wed, Oct 26, 2011 at 5:44 AM, Kyle Moffett <kyle@xxxxxxxxxxxxxxx> wrote:
>> > On Tue, Oct 25, 2011 at 16:51, Erik Faye-Lund <kusmabite@xxxxxxxxx> wrote:
>> >> Thanks for pointing this out, I completely forgot about write re-ordering.
>> >>
>> >> This is indeed a problem. So, shouldn't replacing "mutex->autoinit =
>> >> 0;" with "InterlockedExchange(&mutex->autoinit, 0)" solve the problem?
>> >> InterlockedExchange generates a full memory barrier:
>> >> http://msdn.microsoft.com/en-us/library/windows/desktop/ms683590(v=vs.85).aspx
>> >
>> > No, I'm afraid that won't solve the issue (at least in GCC, not sure about MSVC)
>> >
>> > A write barrier in one thread is only effective if it is paired with a
>> > read barrier in the other thread.
>> >
>> > Since there's no read barrier in the "while(mutex->autoinit != 0)",
>> > you don't have any guaranteed ordering.
>
> Out of curiosity, where could re-ordering be a problem here?  I'm
> thinking probably at "EnterCriticalSection(&mutex->cs)" and the contents
> of "mutex->cs" not being propagated to the waiting thread.  However,
> shouldn't that be a non-problem, as far as compiler reordering goes,
> because it's an external function call and only the address of mutex->cs
> is passed?
>
> The only other cause I could think of is if ordering at the CPU was
> somehow different (it could be if there're no special provisions for
> calling external functions) or if "InterlockedExchange(&mutex->autoinit,
> 0)" wasn't atomic in updating autoinit and doing the memory barrier.
>
> Either way, I couldn't vouch for the safety of the above logic without
> a memory barrier so this question is purely of an academical nature. :)
>
>> > I guess if MSVC assumes that volatile reads imply barriers then it might work...
>>
>> OK, so I should probably do something like this instead?
>>
>> while (InterlockedCompareExchange(&mutex->autoinit, 0, 0) != 0)
>>       ; /* wait for other thread */
>
> Technically, assuming only the updating of "mutex->cs" is in question,
> the ICE should only be required once after exiting the loop...
>
> There's a question of the propagation of the value of "mutex->autoinit"
> itself, but my take is that the memory barrier on the writing thread
> will push out the updated value across all CPUs, thus preventing an
> infinite loop.  The other factors, value caching and loop optimization
> by the compiler, should be prevented by the "volatile" keyword even with
> gcc or MSVC 2003.
>
>> I really appreciate getting some extra eyes on this, thanks.
>> Concurrent programming is not my strong-suit (as this exercise has
>> shown) ;)
>
> So would I. :)

Ok, so here's the race condition:

Thread1				Thread2
				/* Speculative prefetch */
				prefetch(*mutex);

if (mutex->autoinit) {
if (ICE(&mutex->autoinit, -1, 1) != -1) {
/* Now mutex->autoinit == -1 */
pthread_mutex_init(mutex, NULL);
/* This forces writes out to memory */
ICE(&mutex->autoinit, 0, -1);

				if (mutex->autoinit) {} /* false */
				/* No read barrier here */
				EnterCriticalSection(&mutex->cs);
				/* Use cached mutex->cs from earlier */

Even though you forced the memory write to be ordered in Thread 1 you
did not ensure that the read of autoinit occurred before the read of
mutex->cs in Thread 2.  If the first thing that EnterCriticalSection
does is follow a pointer or read a mutex key value in mutex->cs then
won't necessarily get the right answer.

The rule of memory barriers is the ALWAYS come in pairs.  This simple
example guarantees that Thread2 will read "tmp_a" == 1 if it
previously read "tmp_b" == 1, although getting "tmp_a" == 1 and
"tmp_b" != 1 is still possible.

Thread1:
a = 1;
write_barrier();
b = 1;

Thread2:
tmp_b = b;
read_barrier();
tmp_a = a;

I think there's a Documentation/memory-barriers.txt file in the kernel
source code with more helpful info.

Cheers,
Kyle Moffett

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