Am 25.10.2011 17:42, schrieb Erik Faye-Lund:
> On Tue, Oct 25, 2011 at 5:28 PM, Johannes Sixt <j.sixt@xxxxxxxxxxxxx> wrote:
>> Am 10/25/2011 16:55, schrieb Erik Faye-Lund:
>>> +int pthread_mutex_lock(pthread_mutex_t *mutex)
>>> +{
>>> + if (mutex->autoinit) {
>>> + if (InterlockedCompareExchange(&mutex->autoinit, -1, 1) != -1) {
>>> + pthread_mutex_init(mutex, NULL);
>>> + mutex->autoinit = 0;
>>> + } else
>>> + while (mutex->autoinit != 0)
>>> + ; /* wait for other thread */
>>> + }
>>
>> The double-checked locking idiom. Very suspicious. Can you explain why it
>> works in this case? Why are no Interlocked functions needed for the other
>> accesses of autoinit? ("It is volatile" is the wrong answer to this last
>> question, BTW.)
>
> I agree that it should look a bit suspicious; I'm generally skeptical
> whenever I see 'volatile' in threading-code myself. But I think it's
> the right answer in this case. "volatile" means that the compiler
> cannot optimize away accesses, which is sufficient in this case.
No, it is not, and it took me a train ride to see what's wrong. It has
nothing to do with autoinit, but with all the other memory locations
that are written. See here, with pthread_mutex_init() inlined:
if (mutex->autoinit) {
Assume two threads enter this block.
if (InterlockedCompareExchange(&mutex->autoinit, -1, 1) != -1) {
Only one thread, A, say on CPU A, will enter this block.
InitializeCriticalSection(&mutex->cs);
Thread A writes some values. Note that there are no memory barriers
involved here. Not that I know of or that they would be documented.
mutex->autoinit = 0;
And it writes another one. Thread A continues below to contend for the
mutex it just initialized.
} else
Meanwhile, thread B, say on CPU B, spins in this loop:
while (mutex->autoinit != 0)
; /* wait for other thread */
When thread B arrives here, it sees the value of autoinit that thread A
has written above.
HOWEVER, when it continues, there is NO [*] guarantee that it will also
see the values that InitializeCriticalSection() has written, because
there were no memory barriers involved. When it continues, there is a
chance that it calls EnterCriticalSection() with uninitialized values!
}
[*] If you compile this code with MSVC >= 2005, "No guarantee" is not
true, it's exactly the opposite because Microsoft extended the meaning
of 'volatile' to imply a memory barriere. This is *NOT* true for gcc in
general. It may be true for MinGW gcc, but I do not know.
> Basically, the thread that gets the original 1 returned from
> InterlockedCompareExchange is the only one who writes to
> mutex->autoinit. All other threads only read the value, and the
> volatile should make sure they actually do. Since all 32-bit reads and
> writes are atomic on Windows (see
> http://msdn.microsoft.com/en-us/library/windows/desktop/ms684122(v=vs.85).aspx
> "Simple reads and writes to properly-aligned 32-bit variables are
> atomic operations.") and mutex->autoinit is a LONG, this should be
> safe AFAICT. In fact, Windows specifically does not have any
> explicitly atomic writes exactly for this reason.
There is a difference between atomic and coherent: Yes, 32-bit accesses
are atomic, but they are not automatically coherent: A 32-bit value
written by one CPU is not instantly visible on the other CPU. 'volatile'
as per the C lanugage does not add any guarantees that would be of
interest here. OTOH, Microsoft's definition of 'volatile' does.
> The only ways mutex->autoinit can be updated is:
> - InterlockedCompareExchange compares it to 1, finds it's identical
> and inserts -1
> - intialization is done
> Both these updates happens from the same thread.
>
> Yes, details like this should probably go into the commit message ;)
A comment in the function is preferred!
-- Hannes
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