Re: [Boost-users] thread safety of the mutex ctor

From: t. scott urban (scottu_at_[hidden])
Date: 2005-08-18 11:48:56

On Thu, 2005-08-18 at 11:22 +0000, James E Taylor wrote:
> Are you sure there are no guarantees for function scope statics?
> 
> I think the following is thread safe:
> 
> void
> my_func()
> {
>     static pthread_mutex_t m = PTHREAD_MUTEX_INITIALIZER;
>     pthread_mutex_lock(&m);
>     // do something once
>     pthread_mutex_unlock(&m);
> }
> 
> because the C-style PTHREAD_MUTEX_INITIALIZER doesn't involve a constructor call: the static is initialised before _any_ threads are running.
> 
> Also, constructors of global statics are not guaranteed to be single-threaded for the same reason you can't safely reference one static from the constructor of another (The Static Initialisation Order Fiasco); one constructor could create a thread that goes and uses an uninitialised static, whilst the main thread tries to initialise it.
> 
> Do you know whether the mutex ctor is thread-safe?

Well, like I said, I think this is all undefined or at least
implementation dependent.  This is what my pthread docs say about the
above code, though:

man pthread_mutex_init:

Without  static initialization, a self-initializing routine foo() might
look as follows:

     static pthread_once_t foo_once = PTHREAD_ONCE_INIT;
     static pthread_mutex_t foo_mutex;

     void foo_init()
     {
          pthread_mutex_init(&foo_mutex, NULL);
     }

     void foo()
     {
          pthread_once(&foo_once, foo_init);
          pthread_mutex_lock(&foo_mutex);
          /* Do work. */
          pthread_mutex_unlock(&foo_mutex);
     }

With static initialization, the same routine could be coded as follows:

     static pthread_mutex_t foo_mutex = PTHREAD_MUTEX_INITIALIZER;

     void foo()
     {
          pthread_mutex_lock(&foo_mutex);
          /* Do work. */
          pthread_mutex_unlock(&foo_mutex);
     }

....

End quote

-- 
t. scott urban <scottu_at_[hidden]>