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Threads A, B, C in that order, reach synchronized method f() in a single object.
All have the same priority.
B and C are blocked.
A leaves f().
What thread now begins running in f()? Is it always B on the principle of FIFO? Or is the order undetermined?
If C has a higher priority than B, does that guarantee that C will run rather than B?
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The running thread will block when it must wait for some event to occur (response to an IPC request, wait on a mutex, etc.). The blocked thread is removed from the running array, and the highest-priority ready thread that's at the head of its priority's queue is then allowed to run.
Multiple threads accessing shared data simultaneously may lead to a timing dependent error known as data race condition. Data races may be hidden in the code without interfering or harming the program execution until the moment when threads are scheduled in a scenario (the condition) that break the program execution.
Only one thread can hold a lock at a time. If a thread tries to take a lock that is already held by another thread, then it must wait until the lock is released.
Conceptually, it is similar to cooperative multi-tasking used in real-time operating systems, in which tasks voluntarily give up execution time when they need to wait upon some type of the event. This type of multithreading is known as block, cooperative or coarse-grained multithreading.
For all application intents and purposes you can assume the order is totally random. Don't play with priorities - you can easily introduce subtle priority inversion bugs that are very very very hard to catch.
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