Does mutex tell processor not to interrupt thread? - multithreading

I was thinking about this when using mutex.
If you lock the thread with mutex for instance
m.lock();
a += b;
m.unlock();
What happens when the the two threads collide?
Is it like:
1) by locking thread A tells CPU that this must be executed uninterrupted until unlock
and while the a += b statement is happening context switch never occurs.
2) thread A locks, while doing the a += b statement context switch happens, thread B sees that its locked, yelds control and everything is back to A which finishes a += b; operation and unlocks access?
If it's case 2 after all, is there a way to tell CPU that certain parts of the code shall never be interrupted by context switch?

It is 2, and there is no direct way to say that code must run to completion without a context switch. You could probably get close by bumping the execution priority of the thread but such shenanigans are usually frowned upon and suggest a bad design.

Both flavors of mutexes exist, but the first kind is used only in privileged programs (O/S kernel, hardware drivers). User is given only second kind, otherwise a user program could hang the whole system.

Related

linux unnamed posix semaphore sem_destroy, sem_wait method question

when there are waiting semaphores of sem_wait method, I call the sem_destroy method on other thread. But waiting semaphore was not wake up.
In case of mutex, pthread_mutex_destroy was return the value EBUSY when there are some waiting threads.
however sem_destroy return 0 and errno was also set 0.
I want to destroy semaphore after calling sem_destroy to block access as destroyed semaphore and to wake up the waiting thread.
Semaphore handle of Window OS is possible.
please advise me. thank you.
POSIX says this about sem_destroy:
The effect of destroying a semaphore upon which other threads are currently blocked is undefined.
It specifically doesn't say that other threads are woken up. In fact, if sem_t contains a pointer to memory, what it almost certainly does do is free the memory, meaning you then have a use-after-free security problem. (Whether that is the case depends on your libc.)
The general approach of allocation for mutexes and semaphores is that they should be either allocated and freed with their relevant data structure, or they should be allocated before the relevant code needs them and then freed after the entire code is done with them. In C, you cannot safely deallocate data structures (e.g., with sem_destroy) that are in use.
If you want to wake up all users of the semaphore, you must increment it until all users have awoken. You can call sem_getvalue to determine if anyone is waiting on the semaphore and then call sem_post to increment it. Only then can you safely destroy it. Note that this can have a race condition, depending on your code.
However, note that you must be careful that the other code does not continue to use the semaphore after it's destroyed, such as by trying to re-acquire it in a loop. If you are careful to structure your code properly, then you can have confidence that this won't happen.

Kernel Programming - Mutexes

So I'm trying to use mutex_init(), mutex_lock(), mutex_unlock() for thread synchronization.
I am currently trying to schedule threads in a round robin fashion(but more than 1 thread could be running at a time) and I set the current state of a thread to TASK_INTERRUPTIBLE, followed by waking up another thread whose PID, I have in a list.
I need to iterate over this list for my logic.
As I understand it, I need to lock this list as I access its elements, or another thread might miss a new entry while I'm making changes to it. Also, as one mutex has locked a resource, no other mutex can unlock it, until the original mutex releases it.
But, I'm still not sure if I'm locking it correctly. (I release the lock before I call schedule(), and re-lock after that)
I declare a mutex locally within a thread and lock the list. After my current thread locks
mutex_lock(&lock);
and I iterate over the list, till I find something(or ends if it doesn't find anything), then unlocks.
mutex_unlock(&lock);
I assume locking while I iterate is legal. I have never seen examples of this though.
Also, is it normal for the process to have a state of (TASK_UNINTERRUPTIBLE) while it holds a mutex lock?
EDIT : I am adding some more information based on the answer below.
It is possible my program may be run on a virtual machine with a single core. Therefore, I do not want to risk infinite polling using spin_lock().
I am trying to maintain scheduling between threads that have a certain id. For example if there are 4 threads. 2 in set 'A' and 2 in set 'B'. I allow only 1 thread to run in each set. But I switch between threads in a given set. However, a thread in set 'A' should not switch to any thread in set 'B'
(I know the kernel scheduler wont be perfect, so an approximate switching will do).
My Reasoning for TASK_STATE's:
1) Initial thread that gets created is running.
2) If another thread in the same set is running (and this one hasn't executed for a given time). Set other thread to TASK_INTERRUPTIPLE, while calling schedule(); Note: There can be more than 2 threads in each set, but let's keep it simple by considering only 2 for now.
3) If it has executed for enough time, set this task to TASK_INTERRUPTIPLE, set the other task in the same set to TASK_RUNNING, while calling schedule();
All this logic happens while I am accessing certain data structures which are locked by a (now) Global Mutex. I unlock the mutex just before I call schedule(), and instantly re-lock afterward. After my logic part is done, I completely unlock the mutex.
Is there anything fundamentally wrong with the approach?
As I understand it, I need to lock this list as I access its elements
Yes, that is true. But if you use a mutex, you're going to be really sad because a call to lock/unlock is a call to the scheduler. Therefore, calling it from inside the scheduler should result in deadlock. What you need to do depends on if your processor is multi-core or (the mythical) single-core. (Is this a virtual system?) On a single-core processor you can disable interrupts. On a multi-core processor, disabling interrupts is not sufficient (it only disables interrupts for that one core, and another core may still be interrupted). The simplest thing to do on a multi-core is to use a spinlock. Unlike the mutex, both of these locking mechanisms can be unlocked from different threads.
I set the current state of a thread to TASK_INTERRUPTIBLE
Is the thread being taken off the CPU? If so, it's not running, so I suspect that TASK_INTERRUPTIBLE is the wrong state. It would be helpful if you could list the possible states for me or if you could describe what the state is supposed to indicate. Because to me "TASK_INTERRUPTIBLE" sounds like a running task.
I declare a mutex locally within a thread and lock the list
Local mutexes are a red flag! The resource you are locking should be protected by a mutex with the same scope. If the list is global, it should have a global mutex to protect it. Threads that want to use the list must first acquire its mutex. Of course, as I already talked about, you probably want to use a different kind of locking to protect the list of ready-to-run processes.
I assume locking while I iterate is legal
It is perfectly legal (assuming of course that your mutual exclusion scheme is bug-free). In fact, it's required. If another thread were allowed to, for example, remove a node from the list while you were reading it, you could end up dereferencing a deleted node.
Also, is it normal for the process to have a state of TASK_UNINTERRUPTIBLE while it holds a mutex lock?
No, not while it holds the lock if the process is currently running on a CPU. A mutex is available to user code. If holding a mutex made the process uninterruptible, that would mean that a process could hijack the system by simply locking a mutex and never releasing it. Now, you will find that the lock and unlock functions need to be uninterruptible on a single-core processor. However, it doesn't make sense to set the state for the process because it's actually the scheduler that must not be interrupted.

Why is interrupt disabled between spin_lock and spin_unlock in Linux?

I was reading the implementation of Linux semaphores. Due to atomicity, signal and wait (up and down in the source code) use spin locks. Then I saw Linux disabled interrupt in spin_lock_irqsave and reenabled interrupt in spin_unlock. This confused me. In my opinion, there is really no point disabling interrupt within a critical section.
For example, proc A (currently active) acquired the lock, proc B (blocked) is waiting for the lock and proc C is doing some unrelated stuff. It makes perfect sense to context switch to C within the critical section between A and B. Even if C also tries to acquire the lock, since the lock is already locked by A, the result would be C being blocked and A resuming execution.
Therefore, I don't know why Linux decided to disable interrupt within critical sections guarded by spin locks. It probably won't cause any problems but seems like a redundant operation to me.
Allow me to start off with a disclaimer that I am not a Linux expert, so my answer may not be the most accurate. Please point out any flaws and problems that you may find.
Imagine if some shared data is used by various parts of the kernel, including operations such as interrupt handlers that need to be fast and cannot block. Let's say system call foo is currently active and has acquired a lock to use/access shared data bar, and interrupts are not disabled when/before acquiring said lock.
Now a (hardware) interrupt handler, e.g. the keyboard, kicks in and also needs access to bar (hardware interrupts have higher priority than system calls). Since bar is currently being locked by syscall foo, the interrupt handler cannot do anything. Interrupt handlers do need to be fast & not be blocked though, so they just keep spinning while trying to acquire the lock, which would cause a deadlock (i.e. system freeze) since syscall foo never gets a chance to finish and release its lock.
If you disable interrupts before trying to acquire the lock in foo, though, then foo will be able to finish up whatever it's doing and ultimately release the lock (and restore interrupts). Any interrupts trying to come in while foo holds the spinlock will be left on the queue, and will be able to start when the lock is released. This way, you won't run into the problem described above. However, care must also be taken to ensure that the lock for bar is held for as short as possible, so that other higher priority operations can take over whenever required.
The answer is very simple: There is no way for the thread that tries to acquire a lock, to know if the ISR that will interrupt it, will try to acquire the same lock. If that will happen, the ISR will spin forever on that same lock and the system will deadlock.
But what if an interrupt wants to signal a waiting thread ? Or want to test the sempahore value ? The irq disabling is not here to prevent context switch between two process, but to protect from irq. It's all in the comment at the beginning of the file :
/*
* Some notes on the implementation:
*
* The spinlock controls access to the other members of the semaphore.
* down_trylock() and up() can be called from interrupt context, so we
* have to disable interrupts when taking the lock. It turns out various
* parts of the kernel expect to be able to use down() on a semaphore in
* interrupt context when they know it will succeed, so we have to use
* irqsave variants for down(), down_interruptible() and down_killable()
* too.
*
* The ->count variable represents how many more tasks can acquire this
* semaphore. If it's zero, there may be tasks waiting on the wait_list.
*/

Mutex lock: what does "blocking" mean?

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)?
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?
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.
Too late to answer but I may facilitate the understanding. I am talking more from implementation perspective rather than theoretical texts.
The word "blocking" is kind of technical homonym. People may use it for sleeping or mere waiting. The term has to be understood in context of usage.
Blocking means Waiting - Assume on an SMP system a thread B wants to acquire a spinlock held by some other thread A. One of the mechanisms is to disable preemption and keep spinning on the processor unless B gets it. Another mechanism probably, an efficient one, is to allow other threads to use processor, in case B does not gets it in easy attempts. Therefore we schedule out thread B (as preemption is enabled) and give processor to some other thread C. In this case thread B just waits in the scheduler's queue and comes back with its turn. Understand that B is not sleeping just waiting rather passively instead of busy-wait and burning processor cycles. On BSD and Solaris systems there are data-structures like turnstiles to implement this situation.
Blocking means Sleeping - If the thread B had instead made system call like read() waiting data from network socket, it cannot proceed until it gets it. Therefore, some texts casually use term blocking as "... blocked for I/O" or "... in blocking system call". Actually, thread B is rather sleeping. There are specific data-structures known as sleep queues - much like luxury waiting rooms on air-ports :-). The thread will be woken up when OS detects availability of data, much like an attendant of the waiting room.
Blocking means just that. It is blocked. It will not proceed until able. You don't say which language you're using, but most languages/libraries have lock objects where you can "attempt" to take the lock and then carry on and do something different depending on whether you succeeded or not.
But in, for example, Java synchronized blocks, your thread will stall until it is able to acquire the monitor (mutex, lock). The java.util.concurrent.locks.Lock interface describes lock objects which have more flexibility in terms of lock acquisition.

synchronising threads with mutexes

In Qt, I have a method which contains a mutex lock and unlock. The problem is when the mutex is unlock it sometimes take long before the other thread gets the lock back. In other words it seems the same thread can get the lock back(method called in a loop) even though another thread is waiting for it. What can I do about this? One thread is a qthread and the other thread is the main thread.
You can have your thread that just unlocked the mutex relinquish the processor. On Posix, you do that by calling pthread_yield() and on Windows by calling Sleep(0).
That said, there is no guarantee that the thread waiting on the lock will be scheduled before your thread wakes up again.
It shouldn't be possible to release a lock and then get it back if some other thread is already waiting on it.
Check that you actually releasing the lock when you think you do. Check that waiting thread actually waits (and not spins a loop with a trylock tests and sleeps, I actually done that once and was very puzzled at first :)).
Or if waiting thread really never gets time to even reach locking code, try QThread::yieldCurrentThread(). This will stop current thread and give scheduler a chance to give execution to somebody else. Might cause unnecessary switching depending on tightness of your loop.
If you want to make sure that one thread has priority over the other ones, an option is to use a QReadWriteLock. It's adapted to a typical scenario where n threads are going to read a value in a infinite loop, with only one thread updating it. I think it's the scenario you described.
QReadWriteLock offers two ways to lock: lockForRead() and lockForWrite(). The threads depending on the value will use the latter, while the thread updating the value (typically via the GUI) will use the former (lockForWrite()) and will have top priority. You won't need to sleep or yield or whatever.
Example code
Let's say you have a QReadWrite lock; somewhere.
"Reader" thread
forever {
lock.lockForRead();
if (condition) {
do_stuff();
}
lock.unlock();
}
"Writer" thread
// external input (eg. user) changes the thread
lock.lockForWrite(); // will block as soon as the reader lock ends
update_condition();
lock.unlock();

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