Re: C++ Threads, what's the status quo?

From:
Pete Becker <pete@versatilecoding.com>
Newsgroups:
comp.lang.c++.moderated
Date:
10 Jan 2007 10:45:59 -0500
Message-ID:
<DMudnZwrn5UInDjYnZ2dnUVZ_oannZ2d@giganews.com>
Greg Herlihy wrote:

Pete Becker wrote:

Le Chaud Lapin wrote:

The mutex is simpler but slower:

Mutex counter_mutex; // global

void increment_counter ()
{
   counter_mutex.acquire();
   ++count;
   counter_mutex.release();
}


The increment has no visible side effects, so the compiler is not
obliged to sandwich it between the acquire and the release. How do you
ensure that the compiler won't do either of those?


By having a stack-based object release and acquire the lock:

    #include <pthread.h>

    typedef pthread_mutex_t mutex;

    struct StMutex
    {
        StMutex( mutex * m )
            : mMutex( m )
        {
            pthread_mutex_lock( mMutex);
        }

        ~StMutex()
        {
            pthread_mutex_unlock( mMutex);
        }

    private:
        mutex * mMutex;
    };

    int main()
    {
        StMutex lockMe( &gMutex);

        count++;
    }


This doesn't meet the requirement, which you snipped. The requirement
was was to write a function that increments the value of the static
variable and can be called safely from multiple threads. main can't be
called from multiple threads.

Assuming that this was written as a callable function, how does this
force the compiler to sandwich the increment between the lock and the
unlock? To the compiler this looks just like the earlier version: a
function call, an unrelated increment, and another function call. The
unrelated increment can be moved without affecting the meaning of a
conforming C++ program.

Although, any solution that uses a lock, critical section or other form
of blocking synchonization is inferior to one that does not. So,
assuming that an atomic compare-and-swap routine is available, the
lock-free solution beats the others, hands-down:

    static int count;

    // returns true if value was updated

    bool CompareAndSwap( long oldValue, long newValue, long *address);

    int main()
    {
        while (true)
        {
            long n = count;

            if ( CompareAndSwap( n, n+1, &count))
                break;
        }
        ...
    }


There is no guarantee that the compiler won't optimize away 'n' and
simply pick up count's value twice to compute the arguments to
CompareAndSwap, leaving open the possibility that another thread will
modify count between the two accesses.

--

    -- Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com)
Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." (www.petebecker.com/tr1book)

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