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Mutex lock: what does "blocking" mean?

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I've been reading up on multithreading and shared resources access and one of the many (for me) new concepts is the mutex lock. What I can't seem to find out is what is actually happening to the thread that finds a "critical section" is locked. It says in many places that the thread gets "blocked", but what does that mean? Is it suspended, and will it resume when the lock is lifted? Or will it try again in the next iteration of the "run loop"?

The reason I ask, is because I want to have system supplied events (mouse, keyboard, etc.), which (apparantly) are delivered on the main thread, to be handled in a very specific part in the run loop of my secondary thread. So whatever event is delivered, I queue in my own datastructure. Obviously, the datastructure needs a mutex lock because it's being modified by both threads. The missing puzzle-piece is: what happens when an event gets delivered in a function on the main thread, I want to queue it, but the queue is locked? Will the main thread be suspended, or will it just jump over the locked section and go out of scope (losing the event)?

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zmippie Avatar asked Oct 20 '10 21:10

zmippie


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Is mutex lock blocking?

mutex::lock Locks the mutex. If another thread has already locked the mutex, a call to lock will block execution until the lock is acquired.

What happens when a thread is blocked?

A thread moves to the blocked state when it wants to access an object that is being used (locked) by another thread. Once that resource is available (unlocked) for the thread, it is no longer blocked and moves to the runnable state.

Does mutex block the thread?

If the mutex is already locked by another thread, the subroutine blocks the calling thread until the mutex is unlocked. If the mutex is already locked by the calling thread, the subroutine might block forever or return an error depending on the type of mutex.

What happens when mutex lock?

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2 Answers

The simplest way to think of it is that the blocked thread is put in a wait ("sleeping") state until the mutex is released by the thread holding it. At that point the operating system will "wake up" one of the threads waiting on the mutex and let it acquire it and continue. It's as if the OS simply puts the blocked thread on a shelf until it has the thing it needs to continue. Until the OS takes the thread off the shelf, it's not doing anything. The exact implementation -- which thread gets to go next, whether they all get woken up or they're queued -- will depend on your OS and what language/framework you are using.

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tvanfosson Avatar answered Oct 13 '22 21:10

tvanfosson


Blocked means execution gets stuck there; generally, the thread is put to sleep by the system and yields the processor to another thread. When a thread is blocked trying to acquire a mutex, execution resumes when the mutex is released, though the thread might block again if another thread grabs the mutex before it can.

There is generally a try-lock operation that grab the mutex if possible, and if not, will return an error. But you are eventually going to have to move the current event into that queue. Also, if you delay moving the events to the thread where they are handled, the application will become unresponsive regardless.

A queue is actually one case where you can get away with not using a mutex. For example, Mac OS X (and possibly also iOS) provides the OSAtomicEnqueue() and OSAtomicDequeue() functions (see man atomic or <libkern/OSAtomic.h>) that exploit processor-specific atomic operations to avoid using a lock.

But, why not just process the events on the main thread as part of the main run loop?

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Jeremy W. Sherman Avatar answered Oct 13 '22 19:10

Jeremy W. Sherman