winapi - Win32 Named mutex not released when process crashes

Question

I have 2 processes (A, B) sharing the same mutex (using WaitForSingleObject / ReleaseMutex calls). Everything works fine, but when process A crashes, process B is humming along happily. When I restart process A, there's a deadlock.

Deeper investigation reveals that process B can successfully call ReleaseMutex() twice after process A crashes.

My interpretation: After process A crashes, the mutex is still locked, but ownership of the mutex transfers readily to process B (which is a bug). That's why it's humming along happily, calling WaitForSingleObject (getting WAIT_OBJECT_0 in return) and ReleaseMutex (getting TRUE in return).

Is it possible to use a named synchronization primitive similar to Mutex in such a way that a crash in process A will release the mutex?

One solution is to use SEH and catch the crash and release mutex, but I really hope Windows has a robust primitive that doesn't deadlock like that on process crash.

Answer 1

Some basic assumptions you have to make here about how a mutex works on Windows:

• a mutex is an operating system object that's reference-counted. It will not disappear until the last handle on the mutex is closed
• any handle that's left unclosed when a process terminates is closed by the operating system, decrementing the reference count
• a mutex is re-entrant, calling WaitForSingleObject on a mutex on the same thread succeeds and needs to be balanced with an equal number of ReleaseMutex calls
• an owned mutex becomes abandoned when the thread that owns it terminates without calling ReleaseMutex. Calling WaitForSingleObject on a mutex in this state generates the WAIT_ABANDONED error return code
• it is never a bug in the operating system.

So you can draw conclusions from this by what you observed. Nothing happens to the mutex when A crashes, B still has an handle on it. The only possible way B can notice that A crashed is when A crashed while it owned the mutex. Very low odds for that and easily observed since B will deadlock. Far more likely is that B will happily motor on since it is now completely unobstructed, nobody else is going to acquire the mutex anymore.

Furthermore, a deadlock when A starts back proves something you already knew: B owns the mutex permanently for some reason. Possibly because it acquired the mutex recursively. You know this because you noticed you had to call ReleaseMutex twice. This is a bug you need to fix.

You'll need to protect yourself against a crashing sibling process and you need to write explicit code for that. Call OpenProcess on the sibling to obtain a handle on the process object. A WaitForSingleObject call on the handle will complete when the process terminates.