According to the official doc (emphasis mine):
PTHREAD_MUTEX_ROBUST
If the process containing the owning thread of
a robust mutex terminates while holding the mutex lock, the next
thread that acquires the mutex shall be notified about the termination
by the return value [EOWNERDEAD] from the locking function. If the
owning thread of a robust mutex terminates while holding the mutex
lock, the next thread that acquires the mutex may be notified about
the termination by the return value [EOWNERDEAD]...
The doc seems to be differentiating specifically the cases for out-of-process and in-process thread termination and have chosen the wording shall and may carefully. Does this imply that the robustness is mandatory in the out-of-process case, but is optional in the in-process case (which may result in a deadlock)?
Related
Background
I often stumble open cases where the order of lock acquisitions must be the same as real-time order of lock attempts. Those cases are usually about semaphore like locks.
Theory
From what I read in "The Art of Multiprocessor Programming", deadlock freedom and first-come-first-served guarantees are sufficient to make the lock starvation free. Deadlock freedom seems to be on the users since they have to remember to unlock properly. I have looked at possible types of mutexes provided on the pthreads manual page, but it doesn't seem to mention any ordering on lock acquisitions.
Question
Does pthread mutex guarantee starvation freedom? Are there implementations that do (I'm mainly concerned about linux family and macOS)? Are semaphore guaranteed the same properties as mutex?
This is quite a general computer science question and not specific to any OS or framework.
So I am a little confused by the overhead associated with switching tasks on a thread pool. In many cases it doesn't make sense to give every job its own specific thread (we don't want to create too many hardware threads), so instead we put these jobs into tasks which can be scheduled to run on a thread. We setup up a pool of threads and then dynamically allocate the tasks to run on a thread taken from the thread pool.
I am just a little confused (can't find a in depth answer) on the overhead associated with switching tasks on a specific thread (in the thread pool). A DrDobbs article (sourced below) states it does but I need a more in depth answer to what is actually happening (a cite-able source would be fantastic :)).
By definition, SomeWork must be queued up in the pool and then run on
a different thread than the original thread. This means we necessarily
incur queuing overhead plus a context switch just to move the work to
the pool. If we need to communicate an answer back to the original
thread, such as through a message or Future or similar, we will incur
another context switch for that.
Source: http://www.drdobbs.com/parallel/use-thread-pools-correctly-keep-tasks-sh/216500409?pgno=1
What components of the thread are actually switching? The thread itself isn't actually switching, just the data that is specific to the thread. What is the overhead associated with this (more, less or the same)?
let´s clarify first 5 key concepts here and then discuss how they correlates in a thread pool context:
thread:
In a brief resume it can be described as a program execution context, given by the code that is being run, the data in cpu registries and the stack. when a thread is created it is assigned the code that should be executed in that thread context. In each cpu cycle the thread has an instruction to execute and the data in cpu registries and stack in a given state.
task:
Represents a unit of work. It's the code that is assigned to a thread to be executed.
context switch (from wikipedia):
Is the process of storing and restoring the state (context) of a thread so that execution can be resumed from the same point at a later time. This enables multiple processes to share a single CPU and is an essential feature of a multitasking operating system. What constitutes the context is as explained above is the code that is being executed, the cpu registries and the stack.
What is context switched is the thread. A task represents only a peace of work that can be assigned to a thread to be executed. At given moment a thread can be executing a task.
Thread Pool (from wikipedia):
In computer programming, the thread pool is where a number of threads are created to perform a number of tasks, which are usually organized in a queue.
Thread Pool Queue:
Where tasks are placed to be executed by threads in the pool. This data structure is a shared peace of memory where threads may compete to queue/dequeue, may lead to contention in high load scenarios.
Illustrating a thread pool usage scenario:
In your program (eventually running in the main thread), you create a task and schedules it to be executed in thread pool.
The task is queued in the thread pool queue.
When a thread from the pool executes it dequeues a task from the pool and starts to executed it.
If there is no free cpus to execute the thread from the pool, the operating system at some point (depending on thread scheduler policy and thread priorities) will stop a thread from executing, context switching to other thread.
the operating system can stop the execution of a thread at any time, context switching to another thread, returning latter to continue where it stopped.
The overhead of the context switching is augmented when the number of active threads that competes for cpus grows. Thus, ideally, a thread pool tries to use the minimum necessary threads to occupy all available cpus in a machine.
If your tasks haven't code that blocks somewhere, context switching is minimized because it is used no more threads than the available cpus on machine.
Of course if you have only one core, your main thread and the thread pool will compete for the same cpu.
The article probably talks about the case in which work is posted to the pool and the result of it is being waited for. Running a task on the thread-pool in general does not incur any context switching overhead.
Imagine queueing 1000 work items. A thread-pool thread will executed them one after the other. All of that without a single context switch in between.
Switching happens doe to waiting/blocking.
Is there a foolproof way to automatically release mutexes held by a thread when that thread is exiting (in its destructor)?
The approach I have been taking is to create a structure for each mutex which hold the identity of the thread that holds it, and then in the destructor to scan through this list and if any mutexes match the thread being finished, to release it then. But I'm thinking that this actually has a race condition: what happens if after I lock the mutex but before I set the data structure the destructor is called?
I've also looked at pthread_mutexattr_setrobust_np, but my understanding is that np functions are non-portable, and I have had issues with that in the past.
For reference, each thread is associated with a TCP/IP connection, and locking/unlocking occurs in response to requests over this connection. If the connection abnormally closes I need to clean up i.e. release any locks held.
I found a solution which appears to work. First, I use an error checking mutex (PTHREAD_ERRORCHECK_MUTEX_INITIALIZER or PTHREAD_ERRORCHECK_MUTEX_INITIALIZER_NP).
Next, in the destructor, I trying to unlock all mutexes, with the idea being any mutex not owned by the thread will be left alone, but any mutex owned by the thread will be released.
For some reason even mutexes owned by the thread return EPERM, but a subsequent attempt to re-lock the mutex from another thread succeeds whereas without trying to unlock another attempt will deadlock. Conversely, other mutexes not owned by the destructed thread are still found to be locked after the destructor runs.
pthread_mutex_trylock detects deadlocks, doesn't block, then why would you even "need" pthread_mutex_lock?
Perhaps when you deliberately want the thread to block? But in that case it may result in a deadlock?
pthread_mutex_trylock does not detect deadlocks.
You can use it to avoid deadlocks but you have to do that by wrapping your own code around it, effectively multiple calls to pthread_mutex_trylock in a loop with a time-out, after which your thread releases all its resources.
In any case, you can avoid deadlocks even with pthread_mutex_lock if you just follow the simple rule that all threads allocate resources in the same order.
You use pthread_mutex_lock if you just want to efficiently wait until the resource is available, without having to spin on the mutex, something which is often very inefficient. Properly designed multi-threaded applications have no need for the pthread_mutex_trylock variant.
Locks should only be held for the absolute minimum time to do the work and, if that's too long, you can generally redesign things so the lock time is less (such as by using the mutex to only copy data to a thread's local data areas, and having the long-running bit work on that after the mutex is released).
The pseudo-code:
while not pthread_mutex_trylock:
yield
will continue to run your thread, waiting for the lock to be available, especially since there is no pthread_yield() in POSIX threads (though it's sometimes provided as a non-portable extension).
That means, at worst, the code segment above won't even be able to portably yield the CPU, therefore chewing up the rest of it's quantum every time through the scheduler cycle.
And at best, it will still activate the thread once per scheduler cycle just to see if the mutex can be obtained.
Whereas:
pthread_mutex_lock
will most likely totally pause your thread until the lock is made available, since it will move it to a waiting queue until the current lock holder releases the mutex.
That's probably the major reason why you should prefer pthread_mutex_lock to pthread_mutex_trylock.
Perhaps when you deliberately want the thread to block?
Yup, exactly in this case. But you can mimic pthread_mutex_lock() behavior with something like that
while(pthread_mutex_trylock(&mtx))
pthread_yield()
I want to create a lot of threads for a writing into a thread, and after writing I call exit... But, when I call exit do I free up the stack or do I still consume it??
In order to avoid resource leaks, you have to do one of these 2:
Make sure some other thread call pthread_join() on the thread
Create the thread as 'detached', which can either be done by setting the proper pthread attribute to pthread_create, or by calling the pthread_detach() function.
Failure to do so will often result in the entire stack "leaking" in many implementations.
The system allocates underlying storage for each thread, (thread ID, thread retval, stack), and this will remain in the process space (and not be recycled) until the thread has terminated and has been joined by other threads.
If you have a thread which you don't care how the thread terminates, and a detached thread is a good choice.
For detached threads, the system recycles its underlying resources automatically after the thread terminates.
source article: http://www.ibm.com/developerworks/library/l-memory-leaks/