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.
Related
I am writing a program which creates a thread that prints 10 numbers. When it prints 5 of them, it waits and it is notifying the main thread and then it continues for the next 5 numbers
This is test.c
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <time.h>
#include <pthread.h>
#include <unistd.h>
int rem = 10;
int count = 5;
pthread_mutex_t mtx;
pthread_cond_t cond1;
pthread_cond_t cond2;
void *f(void *arg)
{
int a;
srand(time(NULL));
while (rem > 0) {
a = rand() % 100;
printf("%d\n",a);
rem--;
count--;
if (count==0) {
printf("time to wake main thread up\n");
pthread_cond_signal(&cond1);
printf("second thread waits\n");
pthread_cond_wait(&cond2, &mtx);
printf("second thread woke up\n");
}
}
pthread_exit(NULL);
}
int main()
{
pthread_mutex_init(&mtx, 0);
pthread_cond_init(&cond1, 0);
pthread_cond_init(&cond2, 0);
pthread_t tids;
pthread_create(&tids, NULL, f, NULL);
while(1) {
if (count != 0) {
printf("main: waiting\n");
pthread_cond_wait(&cond1, &mtx);
printf("5 numbers are printed\n");
printf("main: waking up\n");
pthread_cond_signal(&cond2);
break;
}
pthread_cond_signal(&cond2);
if (rem == 0) break;
}
pthread_join(tids, NULL);
}
The output of the program is:
main: waiting
//5 random numbers
time to wake main thread up
second thread waits
5 numbers are printed
main: waking up
Since I do pthread_cond_signal(&cond2);I thought that the thread will wake up and prints the rest numbers but this is not the case. Any ideas why? Thanks in advance.
Summary
The issues have been summarized in comments, or at least most of them. So as to put an actual answer on record, however:
Pretty much nothing about the program's use of shared variables and synchronization objects is correct. It's behavior is undefined, and the specific manifestation observed is just one of the more likely in a universe of possible behaviors.
Accessing shared variables
If two different threads access (read or write) the same non-atomic object during their runs, and at least one of the accesses is a write, then all accesses must be properly protected by synchronization actions.
There is a variety of these, too large to cover comprehensively in a StackOverflow answer, but among the most common is to use a mutex to guard access. In this approach, a mutex is created in the program and designated for protecting access to one or more shared variables. Each thread that wants to access one of those variables locks the mutex before doing so. At some later point, the thread unlocks the mutex, lest other threads be permanently blocked from locking the mutex themselves.
Example:
pthread_mutex_t mutex; // must be initialized before use
int shared_variable;
// ...
void *thread_one_function(void *data) {
int rval;
// some work ...
rval = pthread_mutex_lock(&mutex);
// check for and handle lock failure ...
shared_variable++;
// ... maybe other work ...
rval = pthread_mutex_unlock(&mutex);
// check for and handle unlock failure ...
// more work ...
}
In your program, the rem and count variables are both shared between threads, and access to them needs to be synchronized. You already have a mutex, and using it to protect accesses to these variables looks like it would be appropriate.
Using condition variables
Condition variables have that name because they are designed to support a specific thread interaction pattern: that one thread wants to proceed past a certain point only if a certain condition, which depends on actions performed by other threads, is satisfied. Such requirements arise fairly frequently. It is possible to implement this via a busy loop, in which the thread repeatedly tests the condition (with proper synchronization) until it is true, but this is wasteful. Condition variables allow such a thread to instead suspend operation until a time when it makes sense to check the condition again.
The correct usage pattern for a condition variable should be viewed as a modification and specialization of the busy loop:
the thread locks a mutex guarding the data on which the condition is to be computed;
the thread tests the condition;
if the condition is satisfied then this procedure ends;
otherwise, the thread waits on a designated condition variable, specifying the (same) mutex;
when the thread resumes after its wait, it loops back to (2).
Example:
pthread_cond_t cv; // must be initialized before use
void *thread_two_function(void *data) {
int rval;
// some work ...
rval = pthread_mutex_lock(&mutex);
// check for and handle lock failure ...
while (shared_variable < 5) {
rval = pthread_cond_wait(&cv, &mutex);
// check for and handle wait failure ...
}
// ... maybe other work ...
rval = pthread_mutex_unlock(&mutex);
// check for and handle unlock failure ...
// more work ...
}
Note that
the procedure terminates (at (3)) with the thread still holding the mutex locked. The thread has an obligation to unlock it, but sometimes it will want to perform other work under protection of that mutex first.
the mutex is automatically released while the thread is waiting on the CV, and reacquired before the thread returns from the wait. This allows other threads the opportunity to access shared variables protected by the mutex.
it is required that the thread calling pthread_cond_wait() have the specified mutex locked. Otherwise, the call provokes undefined behavior.
this pattern relies on threads to signal or broadcast to the CV at appropriate times to notify any then-waiting other threads that they might want to re-evaluate the condition for which they are waiting. That is not modeled in the examples above.
multiple CVs can use the same mutex.
the same CV can be used in multiple places and with different associated conditions. It makes sense to do this when all the conditions involved are affected by the same or related actions by other threads.
condition variables do not store signals. Only threads that are already blocked waiting for the specified CV are affected by a pthread_cond_signal() or pthread_cond_broadcast() call.
Your program
Your program has multiple problems in this area, among them:
Both threads access shared variables rem and count without synchronization, and some of the accesses are writes. The behavior of the whole program is therefore undefined. Among the common manifestations would be that the threads do not observe each other's updates to those variables, though it's also possible that things seem to work as expected. Or anything else.
Both threads call pthread_cond_wait() without holding the mutex locked. The behavior of the whole program is therefore undefined. "Undefined" means "undefined", but it is plausible that the UB would manifest as, for example, one or both threads failing to return from their wait after the CV is signaled.
Neither thread employs the standard pattern for CV usage. There is no clear associated condition for either one, and the threads definitely don't test one. That leaves an implied condition of "this CV has been signaled", but that is unsafe because it cannot be tested before waiting. In particular, it leaves open this possible chain of events:
The main thread blocks waiting on cond1.
The second thread signals cond1.
The main thread runs all the way at least through signaling cond2 before the second thread proceeds to waiting on cond2.
Once (3) occurs, the program cannot avoid deadlock. The main thread breaks from the loop and tries to join the second thread, meanwhile the second thread reaches its pthread_cond_wait() call and blocks awaiting a signal that will never arrive.
That chain of events can happen even if the issues called out in the previous points is corrected, and it could manifest exactly the observable behavior you report.
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?
Before pthread wait we lock using a mutex so that some other code might not try to change the condition variable. wait then unlocks the mutex and waits for the signal.
Say, in some other thread i had locked the same mutex and after that, i had used 'signal'. and then unlock thread.
when signal is done, the waiting thread wakes up and aquires the mutex again.
Thread1 Thread2
{ {
lock(mutex); lock(mutex);
wait(mutex); signal(mutex);
unlock(mutex); unlock(mutex);
} }
Say the three thread one statements are enclosed in a while(1) loop. Then assume that thread2 locks the mutex, signals it, and unlocks the mutex. and then doesn't end, but goes to sleep.
So will the value of the condition variable be changed permanently? If three statements of thread one are running in infinite lop, will it never wait and just find that the signal has been given? When wait call returns, does it set the value of the condition variable back to initial value?
If yes, can I use create,destroy or initialize methods on the variables to set the value back? If yes, how? What exactly do these functions do?
Thanks,
pthread_cond_signal() will always wake at least one thread that is currently waiting on that condition variable in pthread_cond_wait(). If the same thread or a different thread then calls pthread_cond_wait() again, it will block and wait for another signal.
This means that pthread condition variables must always be paired with some kind of shared data, protected by the mutex that is held when calling pthread_cond_wait(). Before calling pthread_cond_wait(), the thread must check the shared data to see if the condition it wants to wait for has occurred - if not, it shouldn't wait.
The simplest example of such shared data might be a global flag. In your example:
int flag = 0;
Thread 1 {
pthread_mutex_lock(&mutex);
while (!flag)
pthread_cond_wait(&cond, &mutex);
pthread_mutex_unlock(&mutex);
}
Thread 2 {
pthread_mutex_lock(&mutex);
flag = 1;
pthread_mutex_signal(&cond);
pthread_mutex_unlock(&mutex);
}
You can see here that when the condition is "reset" is entirely under your control - for example you could have Thread 1 set flag = 0; before it calls pthread_mutex_unlock().
The shared state is often more complex than a simple flag - for example you might have a producer thread call pthread_mutex_wait() while there is no room in a shared buffer.
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?
I'm implementing a thread with a queue of tasks. As soon as as the first task is added to the queue the thread starts running it.
Should I use pthread condition variable to wake up the thread or there is more appropriate mechanism?
If I call pthread_cond_signal() when the other thread is not blocked by pthread_cond_wait() but rather doing something, what happens? Will the signal be lost?
Semaphores are good if-and-only-if your queue already is thread safe. Also,
some semaphore implementations may be limited by top counter value.
Even it is unlikely you would overrun maximal value.
Simplest and correct way to do this is following:
pthread_mutex_t queue_lock;
pthread_cond_t not_empty;
queue_t queue;
push()
{
pthread_mutex_lock(&queue_lock);
queue.insert(new_job);
pthread_cond_signal(¬_empty)
pthread_mutex_unlock(&queue_lock);
}
pop()
{
pthread_mutex_lock(&queue_lock);
if(queue.empty())
pthread_cond_wait(&queue_lock,¬_empty);
job=quque.pop();
pthread_mutex_unlock(&queue_lock);
}
From the pthread_cond_signal Manual:
The pthread_cond_broadcast() and pthread_cond_signal() functions shall have no effect if there are no threads currently blocked on cond.
I suggest you use Semaphores. Basically, each time a task is inserted in the queue, you "up" the semaphore. The worker thread blocks on the semaphore by "down"'ing it. Since it will be "up"'ed one time for each task, the worker thread will go on as long as there are tasks in the queue. When the queue is empty the semaphore is at 0, and the worker thread blocks until a new task arrives. Semaphores also easily handle the case when more than 1 task arrived while the worker was busy. Notice that you still have to lock access to the queue to keep inserts/removes atomic.
The signal will be lost, but you want the signal to be lost in that case. If there is no thread to wakeup, the signal serves no purpose. (If nobody is waiting for something, nobody needs to be notified when it happens, right?)
With condition variables, lost signals cannot cause a thread to "sleep through a fire". Unless you actually code a thread to go to sleep when there's already a fire, there is no need to "save a signal". When the fire starts, your broadcast will wake up any sleeping threads. And you would have to be pretty daft to code a thread to go to sleep when there's already a fire.
As already suggested, semaphores should be the best choice. If you need a fixed-size queue just use 2 semaphores (as in classical producer-consumer).
In artyom code, it would be better to replace "if" with "while" in pop() function, to handle spurious wakeup.
No effects.
If you check how pthread_condt_signal is implemented, the condt uses several counters to check whether there are any waiting threads to wake up. e.g., glibc-nptl
/* Are there any waiters to be woken? */
if (cond->__data.__total_seq > cond->__data.__wakeup_seq){
...
}