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;
}
Related
I'm reading the book, Modern Operation Systems by AS TANENBAUM and it gives an example explaining condition variable as below. It looks to me there is a deadlock and not sure what I miss.
Lets assume consumer thread starts first. Right after the_mutex is locked, consumer thread is blocked waiting for the condition variable, condc.
If producer is running at this time, the_mutex will still be locked, because consumer never releases it. So producer will also be blocked.
This looks to me a textbook deadlock issue. Did I miss something here? Thx
#include <stdio.h>
#include <pthread.h>
#define MAX 10000000000 /* Numbers to produce */
pthread_mutex_t the_mutex;
pthread_cond_t condc, condp;
int buffer = 0;
void* consumer(void *ptr) {
int i;
for (i = 1; i <= MAX; i++) {
pthread_mutex_lock(&the_mutex); /* lock mutex */
/*thread is blocked waiting for condc */
while (buffer == 0) pthread_cond_wait(&condc, &the_mutex);
buffer = 0;
pthread_cond_signal(&condp);
pthread_mutex_unlock(&the_mutex);
}
pthread_exit(0);
}
void* producer(void *ptr) {
int i;
for (i = 1; i <= MAX; i++) {
pthread_mutex_lock(&the_mutex); /* Lock mutex */
while (buffer != 0) pthread_cond_wait(&condp, &the_mutex);
buffer = i;
pthread_cond_signal(&condc);
pthread_mutex_unlock(&the_mutex);
}
pthread_exit(0);
}
int main(int argc, char **argv) {
pthread_t pro, con;
//Simplified main function, ignores init and destroy for simplicity
// Create the threads
pthread_create(&con, NULL, consumer, NULL);
pthread_create(&pro, NULL, producer, NULL);
}
When you wait on a condition variable, the associated mutex is released for the duration of the wait (that's why you pass the mutex to pthread_cond_wait).
When pthread_cond_wait returns, the mutex is always locked again.
Keeping this in mind, you can follow the logic of the example.
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).
The main question is: How we can wait for a thread in Linux kernel to complete? I have seen a few post concerned about proper way of handling threads in Linux kernel but i'm not sure how we can wait for a single thread in the main thread to be completed (suppose we need the thread[3] be done then proceed):
#include <linux/kernel.h>
#include <linux/string.h>
#include <linux/errno.h>
#include <linux/sched.h>
#include <linux/kthread.h>
#include <linux/slab.h>
void *func(void *arg) {
// doing something
return NULL;
}
int init_module(void) {
struct task_struct* thread[5];
int i;
for(i=0; i<5; i++) {
thread[i] = kthread_run(func, (void*) arg, "Creating thread");
}
return 0;
}
void cleanup_module(void) {
printk("cleaning up!\n");
}
AFAIK there is no equivalent of pthread_join() in kernel. Also, I feel like your pattern (of starting bunch of threads and waiting only for one of them) is not really common in kernel. That being said, there kernel does have few synchronization mechanism that may be used to accomplish your goal.
Note that those mechanisms will not guarantee that the thread finished, they will only let main thread know that they finished doing the work they were supposed to do. It may still take some time to really stop this tread and free all resources.
Semaphores
You can create a locked semaphore, then call down in your main thread. This will put it to sleep. Then you will up this semaphore inside of your thread just before exiting. Something like:
struct semaphore sem;
int func(void *arg) {
struct semaphore *sem = (struct semaphore*)arg; // you could use global instead
// do something
up(sem);
return 0;
}
int init_module(void) {
// some initialization
init_MUTEX_LOCKED(&sem);
kthread_run(&func, (void*) &sem, "Creating thread");
down(&sem); // this will block until thread runs up()
}
This should work but is not the most optimal solution. I mention this as it's a known pattern that is also used in userspace. Semaphores in kernel are designed for cases where it's mostly available and this case has high contention. So a similar mechanism optimized for this case was created.
Completions
You can declare completions using:
struct completion comp;
init_completion(&comp);
or:
DECLARE_COMPLETION(comp);
Then you can use wait_for_completion(&comp); instead of down() to wait in main thread and complete(&comp); instead of up() in your thread.
Here's the full example:
DECLARE_COMPLETION(comp);
struct my_data {
int id;
struct completion *comp;
};
int func(void *arg) {
struct my_data *data = (struct my_data*)arg;
// doing something
if (data->id == 3)
complete(data->comp);
return 0;
}
int init_module(void) {
struct my_data *data[] = kmalloc(sizeof(struct my_data)*N, GFP_KERNEL);
// some initialization
for (int i=0; i<N; i++) {
data[i]->comp = ∁
data[i]->id = i;
kthread_run(func, (void*) data[i], "my_thread%d", i);
}
wait_for_completion(&comp); // this will block until some thread runs complete()
}
Multiple threads
I don't really see why you would start 5 identical threads and only want to wait for 3rd one but of course you could send different data to each thread, with a field describing it's id, and then call up or complete only if this id equals 3. That's shown in the completion example. There are other ways to do this, this is just one of them.
Word of caution
Go read some more about those mechanisms before using any of them. There are some important details I did not write about here. Also those examples are simplified and not tested, they are here just to show the overall idea.
kthread_stop() is a kernel's way for wait thread to end.
Aside from waiting, kthread_stop() also sets should_stop flag for waited thread and wake up it, if needed. It is usefull for threads which repeat some actions infinitely.
As for single-shot tasks, it is usually simpler to use works for them, instead of kthreads.
EDIT:
Note: kthread_stop() can be called only when kthread(task_struct) structure is not freed.
Either thread function should return only after it found kthread_should_stop() return true, or get_task_struct() should be called before start thread (and put_task_struct() should be called after kthread_stop()).
A Naive question ..
I read before saying - "A MUTEX has to be unlocked only by the thread that locked it."
But I have written a program where THREAD1 locks mutexVar and goes for a sleep. Then THREAD2 can directly unlock mutexVar do some operations and return.
==> I know everyone say why I am doing so ?? But my question is - Is this a right behaviour of MUTEX ??
==> Adding the sample code
void *functionC()
{
pthread_mutex_lock( &mutex1 );
counter++;
sleep(10);
printf("Thread01: Counter value: %d\n",counter);
pthread_mutex_unlock( &mutex1 );
}
void *functionD()
{
pthread_mutex_unlock( &mutex1 );
pthread_mutex_lock( &mutex1 );
counter=10;
printf("Counter value: %d\n",counter);
}
int main()
{
int rc1, rc2;
pthread_t thread1, thread2;
if(pthread_mutex_init(&mutex1, NULL))
printf("Error while using pthread_mutex_init\n");
if( (rc1=pthread_create( &thread1, NULL, &functionC, NULL)) )
{
printf("Thread creation failed: %d\n", rc1);
}
if( (rc2=pthread_create( &thread2, NULL, &functionD, NULL)) )
{
printf("Thread creation failed: %d\n", rc2);
}
Pthreads has 3 different kinds of mutexes: Fast mutex, recursive mutex, and error checking mutex. You used a fast mutex which, for performance reasons, will not check for this error. If you use the error checking mutex on Linux you will find you get the results you expect.
Below is a small hack of your program as an example and proof. It locks the mutex in main() and the unlock in the created thread will fail.
#include <stdio.h>
#include <pthread.h>
#include <unistd.h>
#include <errno.h>
#include <stdlib.h>
/*** NOTE THE ATTR INITIALIZER HERE! ***/
pthread_mutex_t mutex1 = PTHREAD_ERRORCHECK_MUTEX_INITIALIZER_NP;
int counter = 0;
void *functionD(void* data)
{
int rc;
if ((rc = pthread_mutex_unlock(&mutex1)) != 0)
{
errno = rc;
perror("other thread unlock result");
exit(1);
}
pthread_mutex_lock(&mutex1);
counter=10;
printf("Thread02: Counter value: %d\n",counter);
return(data);
}
int main(int argc, char *argv[])
{
int rc1;
pthread_t thread1;
if ((rc1 = pthread_mutex_lock(&mutex1)) != 0)
{
errno = rc1;
perror("main lock result");
}
if( (rc1 = pthread_create(&thread1, NULL, &functionD, NULL)))
{
printf("Thread creation failed: %d\n", rc1);
}
pthread_join(thread1, NULL);
}
What you've done is simply not legal, and the behavior is undefined. Mutexes only exclude threads that play by the rules. If you tried to lock mutex1 from thread 2, the thread would be blocked, of course; that's the required thing to do. There's nothing in the spec that says what happens if you try to unlock a mutex you don't own!
A mutex is used to prevent multiple threads from executing code that is only safe for one thread at a time.
To do this a mutex has several features:
A mutex can handle the race conditions associated with multiple threads trying to "lock" the mutex at the same time and always results with one thread winning the race.
Any thread that loses the race gets put to sleep permanently until the mutex is unlocked. The mutex maintains a list of these threads.
A will hand the "lock" to one and only one of the waiting threads when the mutex is unlocked by the thread who was just using it. The mutex will wake that thread.
If that type of pattern is useful for some other purpose then go ahead and use it for a different reason.
Back to your question. Lets say you were protecting some code from multiple thread accesses with a mutex and lets say 5 threads were waiting while thread A was executing the code. If thread B (not one of the ones waiting since they are permanently slept at the moment) unlocks the mutex, another thread will commence executing the code at the same time as thread A. Probably not desired.
Maybe if we knew what you were thinking about using the mutex for we could give a better answer. Are you trying to unlock a mutex after a thread was canceled? Do you have code that can handle 2 threads at a time but not three and there is no mutex that lets 2 threads through at a time?
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;
}