Using Mutexes for interprocess synchronisation - getting AME

Question

I don't understand this behavior:

static Mutex Mut;

    static void Main(string[] args)
    {
        try
        {
            Mut = System.Threading.Mutex.OpenExisting("testmut");
        }
        catch
        {
            Mut = new Mutex(true, "testmut");
        }

        Mut.WaitOne();

        Thread.Sleep(4000);

        Mut.ReleaseMutex();
    }

Starting two instances of this application simultaneously will cause the second application to get an Abandoned Mutex Exception after the first process terminates. Why? I did explicitly release the mutex before terminating there in the first process.

The WaitOne method blocks the current thread until the Mutex has been released. Therefore, after the WaitOne(), you cannot Release it again. — Roy Dictus, Apr 17 '14 at 09:19

score 0 · Answer 1 · answered Apr 17 '14 at 12:55

Okay, the problem was simple. I actually gained Mutex ownership twice in the first process, first by calling Mut = new Mutex(true, "testmut"); (the true flag grants the caller ownership of the Mutex), and then doing a WaitOne(). Obviously, since I only freed it once in the end, I was getting an AM Exception.

Using Mutexes for interprocess synchronisation - getting AME

1 Answers1