Re: [boost] [lockfree] review

Subject: Re: [boost] [lockfree] review
From: Ilya Sokolov (ilyasokol_at_[hidden])
Date: 2011-08-25 07:37:55

On Thu, 25 Aug 2011 02:27:11 +0600, Dave Abrahams <dave_at_[hidden]>  
wrote:
>
> on Wed Aug 24 2011, Alexander Terekhov <terekhov-AT-web.de> wrote:
>
>> Dave Abrahams wrote:
>>>
>>> All I'm saying here is that a C++11 default atomic int is equivalent to
>>> an int, coupled with an associated mutex, where accesses to the int are
>>> always protected by locking the associated mutex.  If you're seriously
>>> disagreeing with *that*, please say explicitly on what grounds.
>>
>>   int i; mutex mi,
>>   int j; mutex mj;
>>
>>   mi.lock();
>>   i = 1;
>>   mi.unlock();
>>
>>   mj.lock();
>>   j = 2;
>>   mj.unlock();
>>
>> can be transformed to
>>
>>   multi_lock(mi, mj); // deadlock free
>>   j = 2;
>>   i = 1;
>>   mi.unlock();
>>   mj.unlock();
>>
>> and thus result in reodering i = 1 and j = 2.
>>
>> With C++11 default atomics (SC) for i and j such reodering is
>> prohibited.
>
> As it is with C++11 mutexes, IIUC.  Another thread that locks the same
> mutexes is not allowed to observe the write to j before the write to i.
> If you have contrary evidence, please point to it.

AFAIK, in the example above another thread can't observe reordering  
without introducing UB.
Here is a more correct example:

int x = 0;
int y = 0;
int r1, r2;
mutex mx, my;

// Thread 1:
mx.lock();
x = 1;
mx.unlock();

my.lock();
r1 = y;
my.unlock();

// Thread 2:
my.lock();
y = 1;
my.unlock();

mx.lock();
r2 = x;
mx.unlock();

Now, it is possible that r1 == r2 == 0, which is not true for the original  
example at
http://www.hpl.hp.com/personal/Hans_Boehm/c++mm/threadsintro.html#introexample

Thread 1
x = 1;
r1 = y;

Thread 2
y = 1;
r2 = x;