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• ContextSwitchDeadlock Was Detected error in C#

  - by assassin

  Hi, I am running a C# application, and during run-time I get the following error: The CLR has been unable to transition from COM context 0x20e480 to COM context 0x20e5f0 for 60 seconds. The thread that owns the destination context/apartment is most likely either doing a non pumping wait or processing a very long running operation without pumping Windows messages. This situation generally has a negative performance impact and may even lead to the application becoming non responsive or memory usage accumulating continually over time. To avoid this problem, all single threaded apartment (STA) threads should use pumping wait primitives (such as CoWaitForMultipleHandles) and routinely pump messages during long running operations. Can anyone please help me out with the problem here? Thanks a lot.

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• Cooperative/Non-preemptive threading avoiding threadlooks?

  - by Wayne

  Any creative ideas to avoid deadlocks on a yield or sleep with cooperative/non-preemptive multitasking without doing an O/S Thread.Sleep(10)? Typically the yield or sleep call will call back into the scheduler to run other tasks. But this can sometime produce deadlocks. Some background: This application has enormous need for speed and, so far, it's extremely fast as compared to other systems in the same industry. One of the speed techniques is cooperative/non-preemptive threading rather then the cost of a context switch from O/S threads. The high level design a priority manager which calls out to tasks depending on priority and processing time. Each task does one "iteration" of work and returns to wait its turn again in the priority queue. The tricky thing with non-preemptive threading is what to do when you want to a particular task to stop in the middle of work and wait for some other event from a different task before continuing. In this case, we have 3 tasks, A B and C where A is a controller that must synchronize the activity of B and C. First, A starts both B and C. Then B yields so C gets invoked. When C yields, A sees they are both inactive, decides it's time for B to run but not time for C yet. Well B is now stuck in a yield that has called C, so it can never run. Sincerely, Wayne

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