Hi there,
I think I have found an error in the mutex routines in Rexxutils under
Linux.
Rexx is: Open Object Rexx Version 4.1.1, Build date: May 21
2012, Addressing Mode: 32
To show the error I created a small script like this:
/* REXX */
sem = "Sem_A"
hdl = SysCreateMutexSem(sem)
if hdl = "" then do
Say "Couldn't create mutex"
hdl = SysOpenMutexSem(sem)
if hdl = 0 then do
Say "Error opening mutex "sem
exit 8
end
end
Say "Mutex handle :"hdl
rc = SysRequestMutexSem(hdl,1)
say "Obained mutex: RC = "rc
pull
rc = SysReleaseMutexSem(hdl)
say "ReleaseMutex: RC = "rc
rc = SysCloseMutexSem(hdl)
say "CloseMutex: RC = "rc
When I start the script it grabs the mutex Sem_A and waits. Now when
starting the script a second time in another shell window
SysCreateMutexSem succeeds which is wrong as the mutex Sem_A does
already exist.
The documentation supports my view: " If you specify name,
SysCreateMutexSem opens the semaphore if the mutex has already been
created."
Having looked into rexxutil.cpp I found that in SysCreateMutexSem a
sem_unlink will be performed for a named pipe.
Please observe that the semaphore semantics are different in Linux
than in Windows.
What happens in our test case when in SysCreateMutexSem the first thing
is a sem_unlink?
The first called rexx script waits after it has gotten the mutex Sem_A.
Now the rexx is called again and in SysCreateMutexSem the call to
sem_unlink leads to the removal of the name Sem_A. The mutex itself
still exists and will die when the first called script stops.
The second called script which is in SysCreatMutexSem now creates a new
mutex with name Sem_A, and thus is able to obtain the mutex.
What do you thin?
--
Manfred
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