Re: [boost] win32 mutex and condition variable algorithms

From: Martin Bonner (martin.bonner_at_[hidden])
Date: 2006-09-19 12:39:40

----Original Message----
From: boost-bounces_at_[hidden]
[mailto:boost-bounces_at_[hidden]] On Behalf Of Anthony Williams
Sent: 19 September 2006 17:02 To: boost_at_[hidden]
Subject: Re: [boost] win32 mutex and condition variable algorithms

> "Martin Bonner" <martin.bonner_at_[hidden]> writes:
> 
>> From: boost-bounces_at_[hidden]
>> [mailto:boost-bounces_at_[hidden]] On Behalf Of Anthony Williams
>>> The mutex is simple, and you can see the code in
>>> boost/thread/win32/basic_mutex.hpp on the thread_rewrite branch.
>>> 
>> 
>> I'm missing something here (I hope).
>> 
>>> The algorithm is:
>>> 
>>> init():
>>>    active_count=0;
>>>    no semaphore
>>> 
>>> lock():
>>>    atomic increment active_count
>> 
>> At this point, thread2 atomically increments active_count as well,
>> so it is now 2. 
>> 
>>>    if new active_count ==1, 
>>>         that's us, so we've got the lock    
>>>    else
>>>         get semaphore, and wait
>> Both threads will go into this branch.  Why don't they wait forever?
> 
> No. The atomic increments are atomic, and return the new value:
                                        ^^^^^^^^^^^^^^^^^^^^^^^^
That's the bit I had missed.  Thanks.

> we don't do an extra read. 
> One thread will see "1", and the other "2".
> 
>>>         now we've got the lock
>>> 
>>> unlock():
>>>    atomic decrement active_count
>>>    if new active_count >0, then other threads are waiting,
>>>        so release semaphore.
>> 
>> 
>> Alternatively,
>> 
>> 	Thread A		Thread B
>> 
>> 	locks mutex
>> 
>> 				atomic increment active_count (in lock)
>> 
>> 	atomic decrements active_count (in unlock)
>> 	new active_count > 0, so other threads are waiting 	release
>> 	semaphore ... oops.  What semaphore?
> 
> Sorry I didn't make it clear. If we go into this branch of unlock, we
> have to call get_semaphore() to get the semaphore to release. This 
> ensures that there is one and only semaphore associated with this 
> mutex, which is returned to all callers, and is therefore released 
> by the unlock().
> 
> Hope that makes things clearer.

It does.  I am now reassured that there are no trivially obvious errors.

... which doesn't mean there aren't any errors :-(

-- 
Martin Bonner
Martin.Bonner_at_[hidden]
Pi Technology, Milton Hall, Ely Road, Milton, Cambridge, CB4 6WZ,
ENGLAND Tel: [phone]