Subj : Re: posix and lock-free algorithms
To   : comp.programming.threads
From : John Doug Reynolds
Date : Thu Aug 11 2005 05:03 am

>  Singleton* Singleton::instance () {
>    Mutex m;
>    if (pInstance == 0) {
>      Lock lock;
>      if (pInstance == 0) {
>        Singleton* tmp = new Singleton;
>        m.lock(); m.unlock(); // just a memory barrier
>        pInstance = tmp;
>
>  While this 'pInstance=tmp;' is executing, another thread could see
> 'pInstance' as zero but not see the Singleton. Your error is in
> thinking that the thread that creates the singleton invoking a
> memory barrier has any effect on other threads that don't themselves
> call memory barrier functions.
>
>  Another thread could see 'pInstance == 0' and not see 'tmp' as
> complete because that thread doesn't call any memory barrier
> functions after testing pInstance but before accessing 'tmp'.
>
>      }
>    }
>    else { m.lock(); m.unlock(); } // just a memory barrier
>    return pInstance;
>  }

But the "reader" thread does place a memory barrier between its read
of pInstance and any attempt to access *pInstance.  If the "first
check" fails (that is, pInstance != 0) in some thread, that thread
immediately executes the second lock-unlock pair.  If the "second
check" fails in some thread, that thread will already have acquired
the Lock and will immediately release it, thus creating a memory
barrier.  So all threads executing this code will encounter a memory
barrier, one way or another.

Doug

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