The problem solvers who create careers with code. LinkedIn YouTube Twitter Facebook Products Finally, the order in happens expect don't assume anything you. Don't assume your thread won't be interrupted at random times. Don't assume the kernel runs threads in the order you started them.

You need to consider the ramifications of external influence on the timing of your various threads, and always assume the worst possible time for things to happen. Parallel programming isn't always parallel, and your program isn't always the only thing running. Not perfect, because there's still a small chance that the test will not test the core feature, but it does prevent deadlock. Check for futex errors and loop until they do what you want them to do, check for the loop starting with the wrong ftx value, and force the main thread to pause for a moment just before loading SS to give the subthread a better chance of being scheduled. Solution: Consider many possible global execution orderings (interleaving of operations). Good thing binary only uses zero and one, yes? The main thread's while loop is waiting for ftx to be nonzero (true), but the subthread hasn't started its loop, so its while loop also sees ftx is 1, and so its loop waits also.ĭeadlock, because two threads don't agree on how to count to two. The main thread has faulted before setting ftx to 2 (due to, for example, a timer tick, peripheral interrupt, or any other outside influence), but has time to recover from the first fault and ftx is still 1 before the subthread is scheduled. What happens if the FUTEX_WAKE fails? How can it fail?Ĭonsider what happens if the subthread is interrupted between the futex call and the while loop, before the second iteration of the main loop. Note the code to synchronize the two threads-the subthread waits for the futex to be 1, and the main thread sets it to 1, using a shared futex. The problem happens when something else in the system happens at just the wrong time. ( Note: the specific operations above are interesting, but not relevant to the problem we're about to discover.) When the kernel next schedules the main thread, it segfaults, calls the handler, and starts the loop all over again.Main thread waits for subthread to set ftx to 0.

Main thread stores a 2 in ftx to start the subthread.The threadproc and main proc are there to switch the LDT back and forth. The general idea of this test is that the modify_ldt syscall might also switch stacks if some of its fields are left uninitialized, avoiding a segfault. * On success, modify_ldt will segfault us synchronously and we will * Make sure the thread is ready after the last test. * If ftx = 2, set it to zero, If ftx = 100, quit. Ready! Go! Set!Ĭonsider this first example, which has been simplified:Ĭhar *p = (char *) malloc(sizeof(char) * 2) In this article, we'll examine two examples of failed concurrency assumptions in the form of two programs from glibc's test suite that weren't operating reliably. Mostly, though, it's because programmers aren't good at writing software where things can happen in any order or at any time. Why? Partly, it's the programming languages, which were designed way back when one processor was the default. Almost every computing device you own today has multiple processors, but pretty much every software written today uses only one at a time.

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