如何确定Linux中是否已存在互斥对象？

Question

I'm currently attempting to port a Multithreading library written for WIN32 to Android/Linux and I seem to have run into a bit of a problem trying to map over a particular feature that WIN32 mutexes have.

From the documentation for Windows Mutex Objects:

Threads in other processes can open a handle to an existing mutex object by specifying the object name in a call to CreateMutex. If a mutex object with that name already exists, GetLastError returns ERROR_ALREADY_EXISTS.

I was not able to find any info on something in the Linux man pages that is similar to what the Windows API offers.

The main reason for my asking this question is that the mutex creation code I'm porting over contains:

isCreator = !(::GetLastError() == ERROR_ALREADY_EXISTS);

where isCreator is a boolean data type. (Not sure what it's used for , but I think its important).

edit:

mutex = PTHREAD_MUTEX_INITIALIZER;
// attributes??

if (pthread_mutex_init(&mutex, NULL) != 0) {
 throw Nv_EXCEPTION(XCPT_ResourceAllocationFailure, GetLastError());
}

Answer 1

Your easiest analog to a named, interprocess mutex is probably a named realtime semaphore initialized to a value of one (1). Try to sem_open("/the_sem", O_CREAT|O_EXCL, mode, 1) . O_EXCL will fail with EEXIST if you aren't the first, and a subsequent sem_open with no flags ought to succeed if you aren't the first:

sem_t *sem;
int isCreator = 0;

if ((sem = sem_open("/the_sem", O_CREAT|O_EXCL, mode, 1)) != SEM_FAILED) {
  // We got here first
  isCreator = 1;
} else {
  if (errno != EEXIST) uh_oh_goodbye();

  // We're not first.  Try again
  sem = sem_open("/the_sem", 0);
  if (sem == SEM_FAILED) uh_oh_goodbye();
}

Note: you can share pthread mutexes (and anonymous realtime semaphores, for that matter) between processes, if the synchronization object is initialized in shared memory. POSIX shared memory realtime extensions permit shared memory to be named , too. There's then a bit of a race between creating the shared memory and initializing the mutex, but in this case attempting to call pthread_mutex_init() should fail with EBUSY for all but the race winner.