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@apptechsys.com>