I'm using pthreads that don't allocate any local variables. For reasons I won't go into here, I need a pthread_cancel() option, and the threads I'm writing should be able to support it (no resources to clean up, OK to stop execution at any point). At the moment, I have a problem because pthread_cancel returns before the pthread is actually finished running, causing problems for shared resources I want to touch only after thread cancellation.
Is there any way I can know when my pthread has well and truly concluded? Is there perhaps a function for this I haven't found, or a parameter I'm not familiar with?
Would
pthread_cancel(thread_handle);
pthread_join(thread_handle, NULL);
do the trick, or is that not guaranteed (since thread_handle may already be invalid)?
I'm pretty new to pthreads, so best practices welcome (beyond "don't use pthread_cancel()," which I've already learned :P ).
The kernel.org manual page is actually doing it. It's safe.
s = pthread_cancel(thr);
if (s != 0)
handle_error_en(s, "pthread_cancel");
/* Join with thread to see what its exit status was */
s = pthread_join(thr, &res);
if (s != 0)
handle_error_en(s, "pthread_join");
Until you call pthread_join on a joinable thread, its tid remains valid. If the thread is joinable (which it must be for pthread_cancel to be safe), then the thread_handle must still be valid.
If the thread was detached, it wouldn't even be safe to call pthread_cancel. What if the thread terminated just as you called it?
Related
I read somewhere that we should lock the mutex before calling pthread_cond_signal and unlock the mutex after calling it:
The pthread_cond_signal() routine is
used to signal (or wake up) another
thread which is waiting on the
condition variable. It should be
called after mutex is locked, and must
unlock mutex in order for
pthread_cond_wait() routine to
complete.
My question is: isn't it OK to call pthread_cond_signal or pthread_cond_broadcast methods without locking the mutex?
If you do not lock the mutex in the codepath that changes the condition and signals, you can lose wakeups. Consider this pair of processes:
Process A:
pthread_mutex_lock(&mutex);
while (condition == FALSE)
pthread_cond_wait(&cond, &mutex);
pthread_mutex_unlock(&mutex);
Process B (incorrect):
condition = TRUE;
pthread_cond_signal(&cond);
Then consider this possible interleaving of instructions, where condition starts out as FALSE:
Process A Process B
pthread_mutex_lock(&mutex);
while (condition == FALSE)
condition = TRUE;
pthread_cond_signal(&cond);
pthread_cond_wait(&cond, &mutex);
The condition is now TRUE, but Process A is stuck waiting on the condition variable - it missed the wakeup signal. If we alter Process B to lock the mutex:
Process B (correct):
pthread_mutex_lock(&mutex);
condition = TRUE;
pthread_cond_signal(&cond);
pthread_mutex_unlock(&mutex);
...then the above cannot occur; the wakeup will never be missed.
(Note that you can actually move the pthread_cond_signal() itself after the pthread_mutex_unlock(), but this can result in less optimal scheduling of threads, and you've necessarily locked the mutex already in this code path due to changing the condition itself).
According to this manual :
The pthread_cond_broadcast() or
pthread_cond_signal() functions
may be called by a thread whether or not it currently owns the mutex that
threads calling pthread_cond_wait()
or pthread_cond_timedwait() have
associated with the condition variable
during their waits; however, if
predictable scheduling behavior is
required, then that mutex shall be
locked by the thread calling
pthread_cond_broadcast() or
pthread_cond_signal().
The meaning of the predictable scheduling behavior statement was explained by Dave Butenhof (author of Programming with POSIX Threads) on comp.programming.threads and is available here.
caf, in your sample code, Process B modifies condition without locking the mutex first. If Process B simply locked the mutex during that modification, and then still unlocked the mutex before calling pthread_cond_signal, there would be no problem --- am I right about that?
I believe intuitively that caf's position is correct: calling pthread_cond_signal without owning the mutex lock is a Bad Idea. But caf's example is not actually evidence in support of this position; it's simply evidence in support of the much weaker (practically self-evident) position that it is a Bad Idea to modify shared state protected by a mutex unless you have locked that mutex first.
Can anyone provide some sample code in which calling pthread_cond_signal followed by pthread_mutex_unlock yields correct behavior, but calling pthread_mutex_unlock followed by pthread_cond_signal yields incorrect behavior?
I have a specific task routine which performs some operations in a specific order, and these operations handle few volatile variables. There is a specific interrupt which updates these volatile variables asynchronously. Hence, the task routine should restart if such an interrupt occurs. Normally FreeRTOS will resume the task, but this will result in wrong derived values, hence the requirement for restarting the routine. I also cannot keep the task routine under critical section, because I should not be missing any interrupts.
Is there a way in FreeRTOS with which I can achieve this? Like a vtaskRestart API. I could have deleted the task and re-created it, but this adds a lot of memory management complications, which I would like to avoid. Currently my only option is to add checks in the routine on a flag to see if a context switch have occured and if yes, restart, else continue.
Googling did not fetch any clue on this. Seems like people never faced such a problem or may be its that this design is poor. In FreeRTOS forum, few who asked for a task-restart didn't seem to have this same problem. stackOverflow didn't have a result on freertos + task + restart. So, this could be the first post with this tag combination ;)
Can someone please tell me if this is directly possible in FreeRTOS?
You can use semaphore for this purpose. If you decide using semaphore, you should do the steps below.
Firstly, you should create a binary semaphore.
The semaphore must be given in the interrupt routine with
xSemaphoreGiveFromISR( Example_xSemaphore, &xHigherPriorityTaskWoken
);
And, you must check taking semaphore in the task.
void vExample_Task( void * pvParameters )
{
for( ;; )
{
if (xSemaphoreTake( Example_xSemaphore, Example_PROCESS_TIME)==pdTRUE)
{
}
}
}
For this purpose you should use a queue and use the queue peek function to yield at your volatile data.
I'm using it as I have a real time timer and this way I make the time available to all my task, without any blocking.
Here it how it goes:
Declare the queue:
xQueueHandle RTC_Time_Queue;
Create the queue of 1 element:
RTC_Time_Queue = xQueueCreate( 1, sizeof(your volatile struct) );
Overwrite the queue everytime your interrupt occurs:
xQueueOverwriteFromISR(RTC_Time_Queue, (void*) &time);
And from other task peek the queue:
xQueuePeek(RTC_GetReadQueue(), (void*) &TheTime, 0);
The 0 at the end of xQueuePeek means you don't want to wait if the queue is empty. The queue peek won't delete the value in the queue so it will be present every time you peek and the code will never stop.
Also you should avoid having variable being accessed from ISR and the RTOS code as you may get unexpected corruption.
I got an EAGAIN when trying to spawn a thread using pthread_create. However, from what I've checked, the threads seem to have been terminated properly.
What determines the OS to give EAGAIN when trying to create a thread using pthread_create? Would it be possible that unclosed sockets/file handles play a part in causing this EAGAIN (i.e they share the same resource space)?
And lastly, is there any tool to check resource usage, or any functions that can be used to see how many pthread objects are active at the time?
Okay, found the answer. Even if pthread_exit or pthread_cancel is called, the parent process still need to call pthread_join to release the pthread ID, which will then become recyclable.
Putting a pthread_join(tid, NULL) in the end did the trick.
edit (was not waitpid, but rather pthread_join)
As a practical matter EAGAIN is almost always related to running out of memory for the process. Often this has to do with the stack size allocated for the thread which you can adjust with pthread_attr_setstacksize(). But there are process limits to how many threads you can run. You can query the hard and soft limits with getrlimit() using RLIMIT_NPROC as the first parameter.
There are quite a few questions here dedicated to keeping track of threads, their number, whether they are dead or alive, etc. Simply put, the easiest way to keep track of them is to do it yourself through some mechanism you code, which can be as simple as incrementing and decrementing a global counter (protected by a mutex) or something more elaborate.
Open sockets or other file descriptors shouldn't cause pthread_create() to fail. If you reached the maximum for descriptors you would have already failed before creating the new thread and the new thread would have already have had to be successfully created to open more of them and thus could not have failed with EAGAIN.
As per my observation if one of the parent process calls pthread_join(), and chilled processes are trying to release the thread by calling pthread_exit() or pthread_cancel() then system is not able to release that thread properly. In that case, if pthread_detach() is call immediately after successful call of pthread_create() then this problem has been solved. A snapshot is here -
err = pthread_create(&(receiveThread), NULL, &receiver, temp);
if (err != 0)
{
MyPrintf("\nCan't create thread Reason : %s\n ",(err==EAGAIN)?"EAGAUIN":(err==EINVAL)?"EINVAL":(err==EPERM)?"EPERM":"UNKNOWN");
free(temp);
}
else
{
threadnumber++;
MyPrintf("Count: %d Thread ID: %u\n",threadnumber,receiveThread);
pthread_detach(receiveThread);
}
Another potential cause: I was getting this problem (EAGAIN on pthread_create) because I had forgotten to call pthread_attr_init on the pthread_attr_t I was trying to initialize my thread with.
I have the following piece of code in thread A, which blocks using pthread_cond_wait()
pthread_mutex_lock(&my_lock);
if ( false == testCondition )
pthread_cond_wait(&my_wait,&my_lock);
pthread_mutex_unlock(&my_lock);
I have the following piece of code in thread B, which signals thread A
pthread_mutex_lock(&my_lock);
testCondition = true;
pthread_cond_signal(&my_wait);
pthread_mutex_unlock(&my_lock);
Provided there are no other threads, would it make any difference if pthread_cond_signal(&my_wait) is moved out of the critical section block as shown below ?
pthread_mutex_lock(&my_lock);
testCondition = true;
pthread_mutex_unlock(&my_lock);
pthread_cond_signal(&my_wait);
My recommendation is typically to keep the pthread_cond_signal() call inside the locked region, but probably not for the reasons you think.
In most cases, it doesn't really matter whether you call pthread_cond_signal() with the lock held or not. Ben is right that some schedulers may force a context switch when the lock is released if there is another thread waiting, so your thread may get switched away before it can call pthread_cond_signal(). On the other hand, some schedulers will run the waiting thread as soon as you call pthread_cond_signal(), so if you call it with the lock held, the waiting thread will wake up and then go right back to sleep (because it's now blocked on the mutex) until the signaling thread unlocks it. The exact behavior is highly implementation-specific and may change between operating system versions, so it isn't anything you can rely on.
But, all of this looks past what should be your primary concern, which is the readability and correctness of your code. You're not likely to see any real-world performance benefit from this kind of micro-optimization (remember the first rule of optimization: profile first, optimize second). However, it's easier to think about the control flow if you know that the set of waiting threads can't change between the point where you set the condition and send the signal. Otherwise, you have to think about things like "what if thread A sets testCondition=TRUE and releases the lock, and then thread B runs and sees that testCondition is true, so it skips the pthread_cond_wait() and goes on to reset testCondition to FALSE, and then finally thread A runs and calls pthread_cond_signal(), which wakes up thread C because thread B wasn't actually waiting, but testCondition isn't true anymore". This is confusing and can lead to hard-to-diagnose race conditions in your code. For that reason, I think it's better to signal with the lock held; that way, you know that setting the condition and sending the signal are atomic with respect to each other.
On a related note, the way you are calling pthread_cond_wait() is incorrect. It's possible (although rare) for pthread_cond_wait() to return without the condition variable actually being signaled, and there are other cases (for example, the race I described above) where a signal could end up awakening a thread even though the condition isn't true. In order to be safe, you need to put the pthread_cond_wait() call inside a while() loop that tests the condition, so that you call back into pthread_cond_wait() if the condition isn't satisfied after you reacquire the lock. In your example it would look like this:
pthread_mutex_lock(&my_lock);
while ( false == testCondition ) {
pthread_cond_wait(&my_wait,&my_lock);
}
pthread_mutex_unlock(&my_lock);
(I also corrected what was probably a typo in your original example, which is the use of my_mutex for the pthread_cond_wait() call instead of my_lock.)
The thread waiting on the condition variable should keep the mutex locked, and the other thread should always signal with the mutex locked. This way, you know the other thread is waiting on the condition when you send the signal. Otherwise, it's possible the waiting thread won't see the condition being signaled and will block indefinitely waiting on it.
Condition variables are typically used like this:
#include <stdio.h>
#include <pthread.h>
#include <unistd.h>
pthread_mutex_t lock = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
int go = 0;
void *threadproc(void *data) {
printf("Sending go signal\n");
pthread_mutex_lock(&lock);
go = 1;
pthread_cond_signal(&cond);
pthread_mutex_unlock(&lock);
}
int main(int argc, char *argv[]) {
pthread_t thread;
pthread_mutex_lock(&lock);
printf("Waiting for signal to go\n");
pthread_create(&thread, NULL, &threadproc, NULL);
while(!go) {
pthread_cond_wait(&cond, &lock);
}
printf("We're allowed to go now!\n");
pthread_mutex_unlock(&lock);
pthread_join(thread, NULL);
return 0;
}
This is valid:
void *threadproc(void *data) {
printf("Sending go signal\n");
go = 1;
pthread_cond_signal(&cond);
}
However, consider what's happening in main
while(!go) {
/* Suppose a long delay happens here, during which the signal is sent */
pthread_cond_wait(&cond, &lock);
}
If the delay described by that comment happens, pthread_cond_wait will be left waiting—possibly forever. This is why you want to signal with the mutex locked.
Both are correct, however for reactivity issues, most schedulers give hand to another thread when a lock is released. I you don't signal before unlocking, your waiting thread A is not in the ready list and thous will not be scheduled until B is scheduled again and call pthread_cond_signal().
The Open Group Base Specifications Issue 7 IEEE Std 1003.1, 2013 Edition (which as far as I can tell is the official pthread specification) says this on the matter:
The pthread_cond_broadcast() or pthread_cond_signal() functions may be
called by a thread whether or not it currently owns the mutex that
threads calling pthread_cond_wait() or pthread_cond_timedwait() have
associated with the condition variable during their waits; however, if
predictable scheduling behavior is required, then that mutex shall be
locked by the thread calling pthread_cond_broadcast() or
pthread_cond_signal().
To add my personal experience, I was working on an application that had code where the conditional variable was destroyed (and the memory containing it freed) by the thread that was woken up. We found that on a multi-core device (an iPad Air 2) the pthread_cond_signal() could actually crash sometimes if it was outside the mutex lock, as the waiter woke up and destroyed the conditional variable before the pthread_cond_signal had completed. This was quite unexpected.
So I would definitely veer towards the 'signal inside the lock' version, it appears to be safer.
Here is nice write up about the conditional variables: Techniques for Improving the Scalability of Applications Using POSIX Thread Condition Variables (look under 'Avoiding the Mutex Contention' section and point 7)
It says that, the second version may have some performance benefits. Because it makes possible for thread with pthread_cond_wait to wait less frequently.
I've been working with pthreads a fair bit recently and there's one little thing I still don't quite get. I know that condition variables are designed to wait for a specific condition to come true (or be 'signalled'). My question is, how does this differ at all from normal mutexes?
From what I understand, aren't condition variables just a mutex with additional logic to unlock another mutex (and lock it again) when the condition becomes true?
Psuedocode example:
mutex mymutex;
condvar mycond;
int somevalue = 0;
onethread()
{
lock(mymutex);
while(somevalue == 0)
cond_wait(mycond, mymutex);
if(somevalue == 0xdeadbeef)
some_func()
unlock(mymutex);
}
otherthread()
{
lock(mymutex);
somevalue = 0xdeadbeef;
cond_signal(mycond);
unlock(mymutex);
}
So cond_wait in this example unlocks mymutex, and then waits for mycond to be signalled.
If this is so, aren't condition variables just mutexes with extra magic? Or do I have a misunderstanding of the fundamental basics of mutexes and condition variables?
The two structures are quite different. A mutex is meant to provide serialised access to a resource of some kind. A condition variable is meant to allow one thread to notify some other thread that some event has occurred.
They aren't exactly mutexes with extra magic, although in some abstractions (monitor as used in java and C#) the condition variable and the mutex are combined into a single unit. The purpose of condition variables is to avoid busing waiting/polling and to hint to the run time which thread(s) should be scheduled "next". Consider how you would write this example without condition variables.
while(1) {
lock(mymutex)
if( somevalue != 0)
break;
unlock(mymutex);
}
if( somevalue == 0xdeadbeef )
myfunc();
You'll be sitting in a tight loop in this thread, burning up a lot of cpu, and making for a lot of lock contention. If locking/unlocking a mutex is cheap enough, you may be in a situation where otherthread never even has a chance to obtain the lock (although real world mutexes generally distinguish between the owning thread and having the lock, as well as having notions of fairness so this is unlikely to happen in reality).
You could reduce the busy waiting by sticking a sleep in,
while(1) {
lock(mymutex)
if( somevalue != 0)
break;
unlock(mymutex);
sleep(1); // let some other thread do work
}
but how long is a good time to sleep for? You'll basically just be guessing. The run time also can't tell why you are sleeping, or what you are waiting on. A condition variable lets the run time be at least somewhat aware of what threads are interested in the same event currently.
The simple answer is that you might want to wake more than one thread from the condition variables, but mutex allows only one thread execute the guarded block.