(oops, didnt send to the list, just Ion. Sorry!) Okay, examples are always the best way to explain things... Should have gone that way from the start :) Receiver code (complete project is attached): std::cout << "Creating or opening semaphore" << std::endl; boost::interprocess::named_semaphore connectSem(boost::interprocess::open_or_create, "semtest-connect",0); std::cout << "Got semaphore, simulating shared memory setup" << "(by sleeping 5 seconds)" << std::endl; sleep(5); std::cout << "Done, allowing senders to connect " << "(by posting semaphore)" << std::endl; connectSem.post(); std::cout << "Continuing with simulated work" << std::endl; for (;;) sleep(1); Sender code (also attached) std::cout << "Creating or opening semaphore" << std::endl; boost::interprocess::named_semaphore connectSem(boost::interprocess::open_or_create, "semtest-connect",0); std::cout << "Waiting on semaphore" << std::endl; connectSem.wait(); std::cout << "Got semaphore, simulating some connection " << "work (by sleeping 5 seconds)" << std::endl; sleep(5); std::cout << "Done, posting semaphore" << std::endl; connectSem.post(); std::cout << "Continuing with simulated work" << std::endl; for (;;) sleep(1); If I start off with no semaphore on disk these programs work fine. It does not matter if I start the sender or the receiver first. If I then kill the programs (and leave the semaphore on disk) it only works as long as I start the receiver before the senders. This would be because the semaphore is left at 1 by the last sender to connect in the first run of the programs. When a sender starts (before the receiver) in a second run it will think that the receiver is already done with its setup, since the wait will return immediately (since the semaphore is at 1). If I try to use named_semaphore::remove on the semaphore before opening_or_creating it (in both sender and receiver) the program will start behaving correctly on windows (since the remove will fail if any process has the semaphore opened), but not on linux (since the semaphore will be unliked and the next process to open_or_create it will then create a completely new semaphore). I've tried to think of a combination of create_only and open_only strategies that would make this possible, but I can't think of one. When I've tried (for example) ACE semaphores I always get a semaphore that is initialized to 0 when an unused one (one that no process has got a handle to) is re-opened. So for the above example when a sender is started with a semaphore on disk (whose last value was 1) the semaphore gets initialized to 0, so the wait will hang and wait for a receiver to post on the semaphore. My tests have been performed on Windows XP with VC80 and Gentoo Linux (kernel 2.6.18-r3) with gcc-4.1.1. One thing that I hope is not confusing the issue is that the Linux system is run inside vmware. I hope I am clearer this time... Cheers Lars Ion Gaztañaga wrote:
Lars Hagström wrote:
Hi,
I am trying to use Boost.Interprocess to get N processes (senders) to send data to one receiver through shared memory. The receiver has to be the one that initializes the shared memory, so no senders can be "let in" before the receiver has completed initialization. I can't guarantee the order that the processes are starting in. To do this I have both the receiver and the senders create_or_open a named_semaphore with 0 as initial value. The senders then immediately do a wait on the semaphore, but the receiver goes ahead and initializes the memory and then does a post on the semaphore. This lets the first sender in, which can then initialize itself and then do a post on the semaphore and start running. The second sender is then let in (because of the post) and it can get going and so on.
I understand.
This approach works very well if the semaphore has been manually deleted from /dev/shm/ or c:\temp\boost_interprocess\, but will not work if there is an old semaphore around.
I understand that the semaphore will still have value 0 and other senders will continue to work fine. Unless a sender crashes in the initialization and can't post the semaphore, of course. In that case, you are lost.
When do you have problems, when launching another receiver if the first one crashes?
I can get around this on Windows by doing a named_semaphore::remove, since that will fail if a process has got it loaded. But this same code will not work on linux, since the semaphore will be unlinked and then all the processes will be able to delete the semaphore and will then be using different semaphores!
I don't understand what you mean here.
I am used to semaphores working so that if it is opened when noone has it loaded it will be initialized to some known value, but this does not appear to be the case with Boost.Interprocess semaphores?
Still lost. In theory, if the semaphore is created you initialize it with a value, if it's already created you get the old value. What do you mean with "loaded"
I'm afraid I need a bit more information to help. I know I am missing something ;-)
Regards,
Ion