Subject: Re: [Boost-users] [thread] variables in thread examples require 'volatile'?
From: Stonewall Ballard (sb.list_at_[hidden])
Date: 2009-12-07 22:16:58
On Dec 7, 2009, at 7:56 PM, Steven Watanabe wrote:
> AMDG
>
> Gabriel Redner wrote:
>> I'm involved in a bit of a dispute in which a colleague is claiming
>> that 'volatile' is necessary to make certain synchronization patterns
>> correct. As part of my response I pointed out that the boost.thread
>> examples don't use it, but he maintains his position - saying that the
>> examples are flawed. I'm frankly somewhat out of my element here, and
>> I'd like to get input from people who have presumably thought about
>> this stuff a good deal.
>>
>> The example in question is here, under "synopsis":
>> http://www.boost.org/doc/libs/1_41_0/doc/html/thread/synchronization.html#thread.synchronization.condvar_ref
>>
>> The claim is that 'volatile' is needed on the boolean, otherwise the
>> compiler might cache the boolean in a register, and the loop would
>> never exit. He cites for support two articles:
>>
>> http://www.ddj.com/cpp/184403766
>> https://www.securecoding.cert.org/confluence/display/seccode/DCL17-C.+Beware+of+miscompiled+volatile-qualified+variables
>>
>> Who's right?
>>
>
> volatile is not needed in that example because access is protected by a mutex.
>
In this example:
----
boost::condition_variable cond;
boost::mutex mut;
bool data_ready;
void process_data();
void wait_for_data_to_process()
{
boost::unique_lock<boost::mutex> lock(mut);
while(!data_ready)
{
cond.wait(lock);
}
process_data();
}
----
What prevents the compiler from leaving data_ready in a register while in the while loop, hence never seeing some other thread setting it? The mutex is released while waiting on the condition, so data_ready is not protected while in cond.wait().
- Stoney
--
Stonewall Ballard
stoney_at_[hidden] http://stoney.sb.org/