This is probably trivial question for some people, but somehow I'm not sure about it.
When waiting with poll for event from kernel, is it that the handling of new event is done in interrupt context ?
If not, does it mean we can sleep/wait (using other commands in handler) in the handler ?
int main (void)
{
struct pollfd fds[2];
int ret;
fds[0].fd = FILENO;
fds[0].events = POLLIN;
fds[1].fd = FILENO;
fds[1].events = POLLOUT;
ret = poll(fds, 2, TIMEOUT * 1000);
if (ret == -1) {
perror ("poll");
return 1;
}
if (!ret) {
return 0;
}
if (fds[0].revents & POLLIN)
{
/********** HANDLING EVENTS HERE ***************/
printf ("FILENO is POLLIN\n");
}
if (fds[1].revents & POLLOUT)
{
/********** HANDLING EVENTS HERE ***************/
printf ("FILENO is POLLOUT\n");
}
return 0;
}
Thank you,
Ran
No (in general).
When you call poll(), the processor context switches to a kernel context, and other processes (and kernel threads) run. Your process will be context switched back in at some point after at least one of your FDs is ready. In general (consider for instance a pipe), interrupt context is not required for this, though note some I/O requires interrupt context to happen (not directly connected to poll()).
Related
I wrote the following code:
void handler (int signal) {
if (signal==SIGVTALRM) {
printf("one second passed\n");
}
if (signal==SIGALRM) {
alarm(1);
//printf("curret context is %ld\n" , thread_tbl[currThreadNum].uc.uc_mcontext.cr2);
currThreadNum=(currThreadNum+1)%THREAD_NUM;
printf("switching from thread #%d to thread #%d\n", ((currThreadNum-1+THREAD_NUM)%THREAD_NUM), currThreadNum);
printf("current thread number is %d\n", currThreadNum);
thread_tbl[(currThreadNum-1+THREAD_NUM)%THREAD_NUM].vtime=+1000;
swapcontext( &(thread_tbl[ (currThreadNum-1+THREAD_NUM)%THREAD_NUM ].uc), &(thread_tbl[currThreadNum ].uc) );
}
}
int ut_start(void) {
int i=0;
struct sigaction sa;
struct itimerval itv;
// set the signal
sa.sa_flags=SA_RESTART;
sigfillset(&sa.sa_mask);
sa.sa_handler = handler;
itv.it_interval.tv_sec=0;
itv.it_interval.tv_usec=100;
itv.it_value=itv.it_interval;
if (sigaction(SIGALRM, &sa, NULL)<0) {
abort();
}
if (sigaction(SIGVTALRM, &sa, NULL)<0) {
abort();
}
setitimer(ITIMER_VIRTUAL, &itv, NULL);
for(i=0; i<TAB_SIZE; i++) {
getcontext(&thread_tbl[i].uc); // get the context the content of current thread
makecontext(&thread_tbl[i].uc, (void(*)(void))func ,1, i); // when this context is activated, func will be executed and then uc.uc_link will get control
}
//start running
alarm(1);
currThreadNum=0;
//printf("currThreadNum=0\n");
swapcontext(&temp_context, &thread_tbl[0].uc);
}
My purpose was to write a program that responds to both SIGVTALRM signal and SIGALRM. However, when I run the program, it seems that the program only react to SIGALRM signals.
In the function ut_start() I started a timer that resets every 100 usec. I don't see any evidence that it works.
One more thing, how can debug this program so I can actually see the timer has started? Is there any variable that I can see in debug mode that tells me something about the status of the timer?
I think I found the answer by myself.
The functions I supplies are only a part of my program. Somewhere in the program, I used the function sleep().
According to the documentation of sleep (http://linux.die.net/man/3/sleep):
sleep() may be implemented using SIGALRM; mixing calls to alarm(2) and
sleep() is a bad idea.Using longjmp(3) from a signal handler or
modifying the handling of SIGALRM while sleeping will cause undefined
results.
When I removed the sleep() function, everything was OK.
I cannot find a proper solution to my problem.
If i have more than one thread in one process. And I want to make only one thread to sleep while running the other threads within the same process, is there any predefined syntax for it or do i have to do my own implementation (sleep) ?
Ideally i want to send a indication from a thread to another thread when it is time for sleep.
Edited (2015-08-24)
I have two main threads, one for sending data over a network, the other receives the data from the network. Beside jitter, the receiving thread does validation and verification and some file management which in time could lead that it will drag behind. What i like to do is to add something like a micro sleep to the sender so that the receiver could catch up. sched_yield() will not help in this case because the HW has a multi core CPU with more than 40 cores.
From your description in the comments, it looks like you're trying to synchronize 2 threads so that one of them doesn't fall behind too far from the other.
If that's the case, you're going about this the wrong way. It is seldom a good idea to do synchronization by sleeping, because the scheduler may incur unpredictable and long delays that cause the other (slow) thread to remain stopped in the run queue without being scheduled. Even if it works most of the time, it's still a race condition, and it's an ugly hack.
Given your use case and constraints, I think you'd be better off using barriers (see pthread_barrier_init(3)). Pthread barriers allow you to create a rendezvous point in the code where threads can catch up.
You call pthread_barrier_init(3) as part of the initialization code, specifying the number of threads that will be synchronized using that barrier. In this case, it's 2.
Then, threads synchronize with others by calling pthread_barrier_wait(3). The call blocks until the number of threads specified in pthread_barrier_init(3) call pthread_barrier_wait(3), at which point every thread that was blocked in pthread_barrier_wait(3) becomes runnable and the cycle begins again. Essentially, barriers create a synchronization point where no one can move forward until everyone arrives. I think this is exactly what you're looking for.
Here's an example that simulates a fast sender thread and a slow receiver thread. They both synchronize with barriers to ensure that the sender does not do any work while the receiver is still processing other requests. The threads synchronize at the end of their work unit, but of course, you can choose where each thread calls pthread_barrier_wait(3), thereby controlling exactly when (and where) threads synchronize.
#include <stdio.h>
#include <pthread.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
pthread_barrier_t barrier;
void *sender_thr(void *arg) {
printf("Entered sender thread\n");
int i;
for (i = 0; i < 10; i++) {
/* Simulate some work (500 ms) */
if (usleep(500000) < 0) {
perror("usleep(3) error");
}
printf("Sender thread synchronizing.\n");
/* Wait for receiver to catch up */
int barrier_res = pthread_barrier_wait(&barrier);
if (barrier_res == PTHREAD_BARRIER_SERIAL_THREAD)
printf("Sender thread was last.\n");
else if (barrier_res == 0)
printf("Sender thread was first.\n");
else
fprintf(stderr, "pthread_barrier_wait(3) error on sender: %s\n", strerror(barrier_res));
}
return NULL;
}
void *receiver_thr(void *arg) {
printf("Entered receiver thread\n");
int i;
for (i = 0; i < 10; i++) {
/* Simulate a lot of work */
if (usleep(2000000) < 0) {
perror("usleep(3) error");
}
printf("Receiver thread synchronizing.\n");
/* Catch up with sender */
int barrier_res = pthread_barrier_wait(&barrier);
if (barrier_res == PTHREAD_BARRIER_SERIAL_THREAD)
printf("Receiver thread was last.\n");
else if (barrier_res == 0)
printf("Receiver thread was first.\n");
else
fprintf(stderr, "pthread_barrier_wait(3) error on receiver: %s\n", strerror(barrier_res));
}
return NULL;
}
int main(void) {
int barrier_res;
if ((barrier_res = pthread_barrier_init(&barrier, NULL, 2)) != 0) {
fprintf(stderr, "pthread_barrier_init(3) error: %s\n", strerror(barrier_res));
exit(EXIT_FAILURE);
}
pthread_t threads[2];
int thread_res;
if ((thread_res = pthread_create(&threads[0], NULL, sender_thr, NULL)) != 0) {
fprintf(stderr, "pthread_create(3) error on sender thread: %s\n", strerror(thread_res));
exit(EXIT_FAILURE);
}
if ((thread_res = pthread_create(&threads[1], NULL, receiver_thr, NULL)) != 0) {
fprintf(stderr, "pthread_create(3) error on receiver thread: %s\n", strerror(thread_res));
exit(EXIT_FAILURE);
}
/* Do some work... */
if ((thread_res = pthread_join(threads[0], NULL)) != 0) {
fprintf(stderr, "pthread_join(3) error on sender thread: %s\n", strerror(thread_res));
exit(EXIT_FAILURE);
}
if ((thread_res = pthread_join(threads[1], NULL)) != 0) {
fprintf(stderr, "pthread_join(3) error on receiver thread: %s\n", strerror(thread_res));
exit(EXIT_FAILURE);
}
if ((barrier_res = pthread_barrier_destroy(&barrier)) != 0) {
fprintf(stderr, "pthread_barrier_destroy(3) error: %s\n", strerror(barrier_res));
exit(EXIT_FAILURE);
}
return 0;
}
Note that, as specified in the manpage for pthread_barrier_wait(3), once the desired number of threads call pthread_barrier_wait(3), the barrier state is reset to the original state that was in use after the last call to pthread_barrier_init(3), which means that the barrier atomically unlocks and resets state, so it is always ready for the next synchronization point, which is wonderful.
Once you're done with the barrier, don't forget to free the associated resources with pthread_barrier_destroy(3).
Using pthreads in linux 2.6.30 I am trying to send a single signal which will cause multiple threads to begin execution. The broadcast seems to only be received by one thread. I have tried both pthread_cond_signal and pthread cond_broadcast and both seem to have the same behavior. For the mutex in pthread_cond_wait, I have tried both common mutexes and separate (local) mutexes with no apparent difference.
worker_thread(void *p)
{
// setup stuff here
printf("Thread %d ready for action \n", p->thread_no);
pthread_cond_wait(p->cond_var, p->mutex);
printf("Thread %d off to work \n", p->thread_no);
// work stuff
}
dispatch_thread(void *p)
{
// setup stuff
printf("Wakeup, everyone ");
pthread_cond_broadcast(p->cond_var);
printf("everyone should be working \n");
// more stuff
}
main()
{
pthread_cond_init(cond_var);
for (i=0; i!=num_cores; i++) {
pthread_create(worker_thread...);
}
pthread_create(dispatch_thread...);
}
Output:
Thread 0 ready for action
Thread 1 ready for action
Thread 2 ready for action
Thread 3 ready for action
Wakeup, everyone
everyone should be working
Thread 0 off to work
What's a good way to send signals to all the threads?
First off, you should have the mutex locked at the point where you call pthread_cond_wait(). It's generally a good idea to hold the mutex when you call pthread_cond_broadcast(), as well.
Second off, you should loop calling pthread_cond_wait() while the wait condition is true. Spurious wakeups can happen, and you must be able to handle them.
Finally, your actual problem: you are signaling all threads, but some of them aren't waiting yet when the signal is sent. Your main thread and dispatch thread are racing your worker threads: if the main thread can launch the dispatch thread, and the dispatch thread can grab the mutex and broadcast on it before the worker threads can, then those worker threads will never wake up.
You need a synchronization point prior to signaling where you wait to signal till all threads are known to be waiting for the signal. That, or you can keep signaling till you know all threads have been woken up.
In this case, you could use the mutex to protect a count of sleeping threads. Each thread grabs the mutex and increments the count. If the count matches the count of worker threads, then it's the last thread to increment the count and so signals on another condition variable sharing the same mutex to the sleeping dispatch thread that all threads are ready. The thread then waits on the original condition, which causes it release the mutex.
If the dispatch thread wasn't sleeping yet when the last worker thread signals on that condition, it will find that the count already matches the desired count and not bother waiting, but immediately broadcast on the shared condition to wake workers, who are now guaranteed to all be sleeping.
Anyway, here's some working source code that fleshes out your sample code and includes my solution:
#include <stdio.h>
#include <pthread.h>
#include <err.h>
static const int num_cores = 8;
struct sync {
pthread_mutex_t *mutex;
pthread_cond_t *cond_var;
int thread_no;
};
static int sleeping_count = 0;
static pthread_cond_t all_sleeping_cond = PTHREAD_COND_INITIALIZER;
void *
worker_thread(void *p_)
{
struct sync *p = p_;
// setup stuff here
pthread_mutex_lock(p->mutex);
printf("Thread %d ready for action \n", p->thread_no);
sleeping_count += 1;
if (sleeping_count >= num_cores) {
/* Last worker to go to sleep. */
pthread_cond_signal(&all_sleeping_cond);
}
int err = pthread_cond_wait(p->cond_var, p->mutex);
if (err) warnc(err, "pthread_cond_wait");
printf("Thread %d off to work \n", p->thread_no);
pthread_mutex_unlock(p->mutex);
// work stuff
return NULL;
}
void *
dispatch_thread(void *p_)
{
struct sync *p = p_;
// setup stuff
pthread_mutex_lock(p->mutex);
while (sleeping_count < num_cores) {
pthread_cond_wait(&all_sleeping_cond, p->mutex);
}
printf("Wakeup, everyone ");
int err = pthread_cond_broadcast(p->cond_var);
if (err) warnc(err, "pthread_cond_broadcast");
printf("everyone should be working \n");
pthread_mutex_unlock(p->mutex);
// more stuff
return NULL;
}
int
main(void)
{
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond_var = PTHREAD_COND_INITIALIZER;
pthread_t worker[num_cores];
struct sync info[num_cores];
for (int i = 0; i < num_cores; i++) {
struct sync *p = &info[i];
p->mutex = &mutex;
p->cond_var = &cond_var;
p->thread_no = i;
pthread_create(&worker[i], NULL, worker_thread, p);
}
pthread_t dispatcher;
struct sync p = {&mutex, &cond_var, num_cores};
pthread_create(&dispatcher, NULL, dispatch_thread, &p);
pthread_exit(NULL);
/* not reached */
return 0;
}
I'm trying to create a thread and let it run until my main signals it to start, which I think is done with SetEvent. But the code in the thread is never executed. Below is the bare code I have stripped down of (I think) unrelated functions. Is the algorithm correct ?
Here is what I thought it did :
When in the main, the thread is created, which means it'll run in the background. When the event is set (SetEvent), the thread picks it up at WaitForSingleObject and then execute the code in the thread, right ?
HANDLE hThread;
HANDLE Event;
DWORD Thread()
{
while(1)
{
wait = WaitForSingleObject(Event, INFINITE)
//This is where I want to execute something
}
}
int _tmain()
{
DWORD dw;
int i;
Event = CreateEvent(NULL,false,false,NULL);
hThread = CreateThread(NULL,0,Thread,EventA,0,NULL);
while(1)
{
if (condition is correct)
{
SetEvent(Event);
}
CloseHandle(Thread);
CloseHandle(Event);
}
return 0;
}
Thanks for having read.
Move CloseHandle lines out of the while loop.
I'm trying to write a simple thread pool program in pthread. However, it seems that pthread_cond_signal doesn't block, which creates a problem. For example, let's say I have a "producer-consumer" program:
pthread_cond_t my_cond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t my_cond_m = PTHREAD_MUTEX_INITIALIZER;
void * liberator(void * arg)
{
// XXX make sure he is ready to be freed
sleep(1);
pthread_mutex_lock(&my_cond_m);
pthread_cond_signal(&my_cond);
pthread_mutex_unlock(&my_cond_m);
return NULL;
}
int main()
{
pthread_t t1;
pthread_create(&t1, NULL, liberator, NULL);
// XXX Don't take too long to get ready. Otherwise I'll miss
// the wake up call forever
//sleep(3);
pthread_mutex_lock(&my_cond_m);
pthread_cond_wait(&my_cond, &my_cond_m);
pthread_mutex_unlock(&my_cond_m);
pthread_join(t1, NULL);
return 0;
}
As described in the two XXX marks, if I take away the sleep calls, then main() may stall because it has missed the wake up call from liberator(). Of course, sleep isn't a very robust way to ensure that either.
In real life situation, this would be a worker thread telling the manager thread that it is ready for work, or the manager thread announcing that new work is available.
How would you do this reliably in pthread?
Elaboration
#Borealid's answer kind of works, but his explanation of the problem could be better. I suggest anyone looking at this question to read the discussion in the comments to understand what's going on.
In particular, I myself would amend his answer and code example like this, to make this clearer. (Since Borealid's original answer, while compiled and worked, confused me a lot)
// In main
pthread_mutex_lock(&my_cond_m);
// If the flag is not set, it means liberator has not
// been run yet. I'll wait for him through pthread's signaling
// mechanism
// If it _is_ set, it means liberator has been run. I'll simply
// skip waiting since I've already synchronized. I don't need to
// use pthread's signaling mechanism
if(!flag) pthread_cond_wait(&my_cond, &my_cond_m);
pthread_mutex_unlock(&my_cond_m);
// In liberator thread
pthread_mutex_lock(&my_cond_m);
// Signal anyone who's sleeping. If no one is sleeping yet,
// they should check this flag which indicates I have already
// sent the signal. This is needed because pthread's signals
// is not like a message queue -- a sent signal is lost if
// nobody's waiting for a condition when it's sent.
// You can think of this flag as a "persistent" signal
flag = 1;
pthread_cond_signal(&my_cond);
pthread_mutex_unlock(&my_cond_m);
Use a synchronization variable.
In main:
pthread_mutex_lock(&my_cond_m);
while (!flag) {
pthread_cond_wait(&my_cond, &my_cond_m);
}
pthread_mutex_unlock(&my_cond_m);
In the thread:
pthread_mutex_lock(&my_cond_m);
flag = 1;
pthread_cond_broadcast(&my_cond);
pthread_mutex_unlock(&my_cond_m);
For a producer-consumer problem, this would be the consumer sleeping when the buffer is empty, and the producer sleeping when it is full. Remember to acquire the lock before accessing the global variable.
I found out the solution here. For me, the tricky bit to understand the problem is that:
Producers and consumers must be able to communicate both ways. Either way is not enough.
This two-way communication can be packed into one pthread condition.
To illustrate, the blog post mentioned above demonstrated that this is actually meaningful and desirable behavior:
pthread_mutex_lock(&cond_mutex);
pthread_cond_broadcast(&cond):
pthread_cond_wait(&cond, &cond_mutex);
pthread_mutex_unlock(&cond_mutex);
The idea is that if both the producers and consumers employ this logic, it will be safe for either of them to be sleeping first, since the each will be able to wake the other role up. Put it in another way, in a typical producer-consumer sceanrio -- if a consumer needs to sleep, it's because a producer needs to wake up, and vice versa. Packing this logic in a single pthread condition makes sense.
Of course, the above code has the unintended behavior that a worker thread will also wake up another sleeping worker thread when it actually just wants to wake the producer. This can be solved by a simple variable check as #Borealid suggested:
while(!work_available) pthread_cond_wait(&cond, &cond_mutex);
Upon a worker broadcast, all worker threads will be awaken, but one-by-one (because of the implicit mutex locking in pthread_cond_wait). Since one of the worker threads will consume the work (setting work_available back to false), when other worker threads awake and actually get to work, the work will be unavailable so the worker will sleep again.
Here's some commented code I tested, for anyone interested:
// gcc -Wall -pthread threads.c -lpthread
#include <stdio.h>
#include <pthread.h>
#include <stdlib.h>
#include <assert.h>
pthread_cond_t my_cond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t my_cond_m = PTHREAD_MUTEX_INITIALIZER;
int * next_work = NULL;
int all_work_done = 0;
void * worker(void * arg)
{
int * my_work = NULL;
while(!all_work_done)
{
pthread_mutex_lock(&my_cond_m);
if(next_work == NULL)
{
// Signal producer to give work
pthread_cond_broadcast(&my_cond);
// Wait for work to arrive
// It is wrapped in a while loop because the condition
// might be triggered by another worker thread intended
// to wake up the producer
while(!next_work && !all_work_done)
pthread_cond_wait(&my_cond, &my_cond_m);
}
// Work has arrived, cache it locally so producer can
// put in next work ASAP
my_work = next_work;
next_work = NULL;
pthread_mutex_unlock(&my_cond_m);
if(my_work)
{
printf("Worker %d consuming work: %d\n", (int)(pthread_self() % 100), *my_work);
free(my_work);
}
}
return NULL;
}
int * create_work()
{
int * ret = (int *)malloc(sizeof(int));
assert(ret);
*ret = rand() % 100;
return ret;
}
void * producer(void * arg)
{
int i;
for(i = 0; i < 10; i++)
{
pthread_mutex_lock(&my_cond_m);
while(next_work != NULL)
{
// There's still work, signal a worker to pick it up
pthread_cond_broadcast(&my_cond);
// Wait for work to be picked up
pthread_cond_wait(&my_cond, &my_cond_m);
}
// No work is available now, let's put work on the queue
next_work = create_work();
printf("Producer: Created work %d\n", *next_work);
pthread_mutex_unlock(&my_cond_m);
}
// Some workers might still be waiting, release them
pthread_cond_broadcast(&my_cond);
all_work_done = 1;
return NULL;
}
int main()
{
pthread_t t1, t2, t3, t4;
pthread_create(&t1, NULL, worker, NULL);
pthread_create(&t2, NULL, worker, NULL);
pthread_create(&t3, NULL, worker, NULL);
pthread_create(&t4, NULL, worker, NULL);
producer(NULL);
pthread_join(t1, NULL);
pthread_join(t2, NULL);
pthread_join(t3, NULL);
pthread_join(t4, NULL);
return 0;
}