PThread Question - c++

I am trying to make a small thread example. I want to have a variable and each thread try to increment it and then stop once it gets to a certain point. Whenever the variable is locked, I want some sort of message to be printed out like "thread x trying to lock, but cannot" so that I KNOW it's working correctly. This is my first day coding threads so feel free to point out anything unnecessary in the code here -
#include <iostream>
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
#define NUM_THREADS 2
pthread_t threads[NUM_THREADS];
pthread_mutex_t mutexsum;
int NUMBER = 0;
void* increaseByHundred(void* threadid) {
if(pthread_mutex_lock(&mutexsum))
cout<<"\nTHREAD "<<(int)threadid<<" TRYING TO LOCK BUT CANNOT";
else {
for(int i=0;i<100;i++) {
NUMBER++;
cout<<"\nNUMBER: "<<NUMBER;
}
pthread_mutex_unlock(&mutexsum);
pthread_exit((void*)0);
}
}
int main(int argc, char** argv) {
int rc;
int rc1;
void* status;
pthread_attr_t attr;
pthread_mutex_init(&mutexsum, NULL);
pthread_attr_init(&attr);
pthread_attr_setdetachstate(&attr, PTHREAD_CREATE_JOINABLE);
rc = pthread_create(&threads[0], &attr, increaseByHundred, (void*)0);
rc1 = pthread_create(&threads[1], &attr, increaseByHundred, (void*)1);
pthread_attr_destroy(&attr);
while(NUMBER < 400)
pthread_join(threads[0], &status);
pthread_mutex_destroy(&mutexsum);
pthread_exit(NULL);
}
I was following a tutorial found here https://computing.llnl.gov/tutorials...reatingThreads
and tried to adapt their mutex example to this idea. The code increments it up to 199 and then stops. I'm guessing because the threads are only doing their routine once. Is there a way make them just do their routine other than when you create them so I could say
while something
do your routine
?
I have the pthread_join there just because it was similar to what that tutorial had on theirs. I don't really even get it that clearly though. I'm pretty sure that line is the problem...I just don't know how to fix it. Any help is appreciated.

Whenever the variable is locked, I want some sort of message to be printed out like "thread x trying to lock, but cannot" so that I KNOW it's working correctly.
Why do you want that? You are just learning about threads. Learn the basics first. Don't go diving off the deep end into pthread_mutex_trylock or mutexes configured for error checking. You need to learn to walk before you can learn how to run.
The basics involves a mutex initialized use with default settings and using pthread_mutex_lock to grab the lock. With the default settings, pthread_mutex_lock will only return non-zero if there are big, big problems. There are only two problems that can occur here: Deadlock, and a bad mutex pointer. There is no recovery from either; the only real solution is to fix the code. About the only thing you can do here is to throw an exception that you don't catch, call exit() or abort(), etc.
That some other thread has locked the mutex is not a big problem. It is not a problem at all. pthread_mutex_lock will block (e.g., go to sleep) until the lock becomes available. A zero return from pthread_mutex_lock means that the calling thread now has the lock. Just make sure you release the lock when you are done working with the protected memory.
Edit
Here's a suggestion that will let you see that the threading mechanism is working as advertised.
Upon entry to increaseByHundred print a time-stamped message indicating entry to the function. You probably want to use C printf here rather than C++ I/O. printf() and related functions are thread-safe. C++ 2003 I/O is not.
After a successful return from pthread_mutex_lock print another time-stamped message indicating that a successful lock.
sleep() for a few seconds and then print yet another time-stamped message prior to calling pthread_mutex_unlock().
Do the same before calling pthread_exit().
One last comment: You are checking for an error return from pthread_mutex_lock. For completeness, and because every good programmer is paranoid as all get out, you should also check the return status from pthread_mutex_unlock.
What about pthread_exit? It doesn't have a return status. You could print some message after calling pthread_exit, but you will only reach that statement if you are using a non-compliant version of the threads library. The function pthread_exit() cannot return to the calling function. Period. Worrying about what happens when pthreads_exit() returns is a tinfoil hat exercise. While good programmers should be paranoid beyond all get out, they should not be paranoid schizophrenic.

pthread_mutex_lock will normally just block until it acquire the lock, and that's why the line cout<<"\nTHREAD "<<(int)threadid<<" TRYING TO LOCK BUT CANNOT"; is not ran.
You also have problems in
while(NUMBER < 400)
pthread_join(threads[0], &status);
because you just have 2 threads and number will never reach 400. You also want to join thread[0] on first iteration, then thread[1]...

pthread_mutex_trylock():
if (pthread_mutex_trylock(&mutex) == EBUSY) {
cout << "OMG NO WAY ITS LOCKED" << endl;
}
It is also worth noting that if the mutex is not locked, it will be able to acquire the lock and then it will behave like a regular pthread_mutex_lock().

Related

Cancelling boost thread from another

Is there a way to cancel a boost::thread from another as in the following?:
boost::thread* thread1(0);
boost::thread* thread2(0);
thread2 = new boost::thread([&](){
//some expensive computation that can't be modified
if(thread1)
thread1->interrupt();
});
thread1 = new boost::thread([&]() {
//some other expensive computation that can't be modified
if(thread2)
thread2->interrupt();
});
thread1->join();
thread2->join();
delete thread1;
delete thread2;
Right now both expensive computations finish without being interrupted. I had figured the joins would be treated as an interruption point, and the main thread would continue after one of the two expensive computations completed.
In general, there is no portable way for one thread to terminate another, without cooperation from the thread being terminated. This question comes up once in a while, it seems (see here and here - although your question is not an exact duplicate).
Barring cooperation from the thread being interrupted (which would have to perform seppuku on notification), if you would like the main thread to continue after the first of the threads has terminated, you could make a condition that each of the child threads fires when it ends.
At this point, you could either let the other thread continue running (possibly detaching it), or just terminate everything.
A non-portable solution for POSIX-compliant systems (e.g. Linux) would be to use pthread_cancel() and then pthread_join() on the Boost thread's native_handle() member, which is of type pthread_t (again, only on POSIX-compliant systems. I can't speak for other systems, like Windows).
Also, you must use a boost::scoped_thread instead of just a boost::thread so that you can "override" (not in the OO-sense) the join/detach behavior that Boost will do when the thread is destroyed. This is necessary because when you call pthread_cancel then pthread_join on a boost::thread, the boost::thread object is still 'joinable' (i.e. boost::thread::joinable() returns true), and so the destructor will exhibit undefined behavior, per the documentation.
With all that being said, if a platform-dependent solution for cancelling threads like this is necessary in your application, I'm not sure there's much to be gained from using boost::threads over plain-old pthreads; still, I suppose there may be a use case for this.
Here's a code sample:
// compilation: g++ -pthread -I/path/to/boost/include -L/path/to/boost/libs -lboost_thread main.cpp
#include <cstdio>
#include <pthread.h>
#include <boost/thread/scoped_thread.hpp>
typedef struct pthreadCancelAndJoin
{
void operator()(boost::thread& t)
{
pthread_t pthreadId = t.native_handle();
int status = pthread_cancel(pthreadId);
printf("Cancelled thread %lu: got return value %d\n", pthreadId, status);
void* threadExitStatus;
status = pthread_join(pthreadId, &threadExitStatus);
printf("Joined thread %lu: got return value %d, thread exit status %ld\n",
pthreadId, status, (long)threadExitStatus);
}
} pthreadCancelAndJoin;
void foo()
{
printf("entering foo\n");
for(int i = 0; i < 2147483647; i++) printf("f"); // here's your 'expensive computation'
for(int i = 0; i < 2147483647; i++) printf("a");
printf("foo: done working\n"); // this won't execute
}
int main(int argc, char **argv)
{
boost::scoped_thread<pthreadCancelAndJoin> t1(foo);
pthread_t t1_pthread = t1.native_handle();
sleep(1); // give the thread time to get into its 'expensive computation';
// otherwise it'll likely be cancelled right away
// now, once main returns and t1's destructor is called, the pthreadCancelAndJoin
// functor object will be called, and so the underlying p_thread will be cancelled
// and joined
return 0;
}
pthread_cancel() will cancel your thread when it reaches a "cancellation point" (assuming the cancel type and cancel state are at their default values, which is the case for boost::thread objects); see the pthreads man page for a list of all cancellation points. You'll notice that those cancellation points include many of the more common system calls, like write, read, sleep, send, recv, wait, etc.
If your 'expensive computation' includes any of those calls down at its lowest level (e.g. in the code sample, printf eventually calls write), it will be cancelled.
Best of all, Valgrind reports no memory leaks or memory errors with this solution.
Finally, a note about your misconception in your question:
I had figured the joins would be treated as an interruption point...
join, or any of the boost::thread interruption functions, for that matter, is only treated as an interruption point for the thread that calls it. Since your main thread is calling join(), the main thread is the thread that experiences the interruption point, not the thread that it is trying to join. E.g. if you call thread1.interrupt() in some thread and then thread1 calls thread2.join(), then thread1 is the one that gets interrupted.

C++ pthread'ed process running slower than single thread issue

I was trying to run a function on multiple pthreads in order to increase efficiency and runtime. This function performs a lot of matrix calculations and print statements. However, when I ran tests in order see the performance improvement, the single threaded code ran faster.
My tests went as follows:
-For the single-threaded: Run a for loop 1:1000 that called the function.
-For the multi-pthreaded: Spawn 100 pthreads, have a queue of 1000 items and a pthread_cond_wait and have the threads run the function until the queue is empty.
Here is my code for the pthreads (single-threaded is just a for loop instead):
# include <iostream>
# include <string>
# include <pthread.h>
# include <queue>
using namespace std;
# define NUM_THREADS 100
int main ( );
queue<int> testQueue;
void *playQueue(void* arg);
void matrix_exponential_test01 ( );
void matrix_exponential_test02 ( );
pthread_mutex_t queueLock;
pthread_cond_t queue_cv;
int main()
{
pthread_t threads[NUM_THREADS];
pthread_mutex_init(&queueLock, NULL);
pthread_cond_init (&queue_cv, NULL);
for( int i=0; i < NUM_THREADS; i++ )
{
pthread_create(&threads[i], NULL, playQueue, (void*)NULL);
}
pthread_mutex_lock (&queueLock);
for(int z=0; z<1000; z++)
{
testQueue.push(1);
pthread_cond_signal(&queue_cv);
}
pthread_mutex_unlock (&queueLock);
pthread_mutex_destroy(&queueLock);
pthread_cond_destroy(&queue_cv);
pthread_cancel(NULL);*/
return 0;
}
void* playQueue(void* arg)
{
bool accept;
while(true)
{
pthread_cond_wait(&queue_cv, &queueLock);
accept = false;
if(!testQueue.empty())
{
testQueue.pop();
accept = true;
}
pthread_mutex_unlock (&queueLock);
if(accept)
{
runtest();
}
}
pthread_exit(NULL);
}
My intuition tells me that the multi-threaded version should run faster, but it doesnt. Is there a reason, or is my code faulty? I am using C++ on Windows, and had to download a library to use pthreads.
First, your code is written in a way that only one thread will run at any time (your mutex is locked the whole time the thread is doing work). So at best you can expect your Code to be as fast as the single threaded version.
Also, all threads reading and writing the same memory each time. This way you force your CPU cores to synchronize their caches, meaning actually more load on the bus than would be caused by a single thread. Since you are not doing any computationally expensive stuff, it is likely that memory bandwidth is your actual bottleneck and thus the bus load added by cache synchronization slows down your program. Take a look at http://en.wikipedia.org/wiki/False_sharing for more information.
If runtest() is CPU-bound -- that is, doesn't do anything which might block on i/o or the like -- then there's not much point starting 100 threads, unless you have 100 cpus/cores ! [Edit: I now notice that runtest() does some print statements... file i/o probably won't block... so won't release the CPU.]
The code as currently shown holds the mutex while filling the queue, so nothing will start until the queue is full. By the time filling of the queue has finished signalling 1000 times, if any have reached the pthread_cond_wait(), then hopefully they will all have been started -- so all 100 will be waiting on the mutex.
As currently shown the waiting in playQueue() is broken. It should be something along the lines of:
pthread_mutex_wait(&queueLock) ;
while (testQueue.empty)
pthread_cond_wait(&queue_cv, &queueLock) ;
if (testQueue.eof)
val = NULL ;
else
val = testQueue.pop ;
pthread_mutex_unlock(&queueLock) ;
But, even when this is all sorted out, there is no guarantee you will see an improvement in performance, unless runtest() does a serious amount of work. [Edit: I now notice that it does "a lot of matrix calculations", which sounds like it could be plenty of work.]
One small suggestion, starting the worker threads and filling the queue could be overlapped, by (for instance) starting one worker with the job of starting all the others, or starting a thread to fill the queue.
Without knowing more about the problem, if the work can be statically divided across the worker threads -- for instance, give the first thread items 0..9, the second thread items 10..19, and so on -- so each worker can ignore all the others, reducing the amount of synchronisation operations.
In addition to the other good answers, you said your runtest() function does I/O.
So you could well be I/O bound, in which case all your threads have to wait in line like everybody else to empty out their buffers.

Thread ending unexpectedly. c++

I'm trying to get a hold on pthreads. I see some people also have unexpected pthread behavior, but none of the questions seemed to be answered.
The following piece of code should create two threads, one which relies on the other. I read that each thread will create variables within their stack (can't be shared between threads) and using a global pointer is a way to have threads share a value. One thread should print it's current iteration, while another thread sleeps for 10 seconds. Ultimately one would expect 10 iterations. Using break points, it seems the script just dies at
while (*pointham != "cheese"){
It could also be I'm not properly utilizing code blocks debug functionality. Any pointers (har har har) would be helpful.
#include <iostream>
#include <cstdlib>
#include <pthread.h>
#include <unistd.h>
#include <string>
using namespace std;
string hamburger = "null";
string * pointham = &hamburger;
void *wait(void *)
{
int i {0};
while (*pointham != "cheese"){
sleep (1);
i++;
cout << "Waiting on that cheese " << i;
}
pthread_exit(NULL);
}
void *cheese(void *)
{
cout << "Bout to sleep then get that cheese";
sleep (10);
*pointham = "cheese";
pthread_exit(NULL);
}
int main()
{
pthread_t threads[2];
pthread_create(&threads[0], NULL, cheese, NULL);
pthread_create(&threads[1], NULL, wait, NULL);
return 0;
}
The problem is that you start your threads, then exit the process (thereby killing your threads). You have to wait for your threads to exit, preferably with the pthread_join function.
If you don't want to have to join all your threads, you can call pthread_exit() in the main thread instead of returning from main().
But note the BUGS section from the manpage:
Currently, there are limitations in the kernel implementation logic for
wait(2)ing on a stopped thread group with a dead thread group leader.
This can manifest in problems such as a locked terminal if a stop sig‐
nal is sent to a foreground process whose thread group leader has
already called pthread_exit().
According to this tutorial:
If main() finishes before the threads it has created, and exits with pthread_exit(), the other threads will continue to execute. Otherwise, they will be automatically terminated when main() finishes.
So, you shouldn't end the main function with the statement return 0;. But you should use pthread_exit(NULL); instead.
If this doesn't work with you, you may need to learn about joining threads here.

Threading in C++ to keep two functions running parallely

I have a code congaing two functions func1 and func2. Role of both the function is same. Keep reading a directory continuously and write the names of file present in their respective log files. Both functions are referring a common log function to write the logs. I want to use introduce threading in my code such that both of them keep on running parallely but both should not access the log function at same time. How to achieve that?
This is a classic case of needing a mutex.
void WriteToLog(const char *msg)
{
acquire(mutex);
logfile << msg << endl;
release(mutex);
}
The above code won't "copy and paste" into your system, since mutexes are system specific - pthread_mutex would be the choice if you are using pthreads. C++11 has it's own mutex and thread functionality, and Windows has another variant.
From Sajal's comments:
tried pthread_create(&thread1, NULL, start_opca, &opca); pthread_join( thread1, NULL); pthread_create(&thread2, NULL, start_ggca, &ggca); pthread_join( thread2, NULL);
But the problem with this is that it will wait for one thread to finish before starting next. I don't want that.
the join function blocks the calling thread, until the thread you call join for, finishes. In your case, calling join on the first thread before creating the second, guarantees that the first thread will end before the second one begins.
You should create the two threads first, then join them both (instead of interspersing the creations and join of both).
Additionally, the access to the log should be extracted into common code for both (a logging function, a logging class etc. Within the extracted code, the log access should be guarded using a mutex.
If you have an implementation (partially) supporting c++11, you should use std::thread and std::mutex for this. Otherwise, you should use boost::thread. If you have access to neither, use pthreads under linux.
On linux, you will need to use pthreads
Since both threads are reading/writing from/to I/O (reading dirs and writing log files) there's no need for multi-threading: you gain no speed improvement parallelizing the task since every I/O access is enqueued at lower levels.
This C language Code may give you some hint. To answer your question:
You should use mutex in pthread to make sure that the log file could only be access by one thread at the same time.
#include <pthread.h>
#include <stdio.h>
pthread_mutex_t LogLock = PTHREAD_MUTEX_INITIALIZER;
char* LogFileName= "test.log";
void* func_tid0( void* a) {
int i;
for(i=0; i < 50; i++ ) {
pthread_mutex_lock(&LogLock);
fprintf((FILE*)a, "write to log by thread0:%d\n", i);
pthread_mutex_unlock(&LogLock);
}
}
void* func_tid1(void* a) {
int i;
for(i=0; i < 50; i++ ) {
pthread_mutex_lock(&LogLock);
fprintf((FILE*)a, "write to log by thread1:%d\n", i);
pthread_mutex_unlock(&LogLock);
}
}
int main() {
pthread_t tid0, tid1;
FILE* fp=fopen(LogFileName, "wb+");
pthread_create(&tid0, NULL, func_tid0, (void*) fp );
pthread_create(&tid1, NULL, func_tid1, (void*) fp );
void* ret;
pthread_join(tid0, &ret);
pthread_join(tid1, &ret);
}
Your another question isn't exist.
Because the main thread is suspend at your first pthread_join, but it's not mean the second thread doesn't run. Actually the second thread is beginning at pthread_create(thread1).
And actually pthread_mutex casuses your program serial.

when to use mutex

Here is the thing: there is a float array float bucket[5] and 2 threads, say thread1 and thread2.
Thread1 is in charge of tanking up the bucket, assigning each element in bucket a random number. When the bucket is tanked up, thread2 will access bucket and read its elements.
Here is how I do the job:
float bucket[5];
pthread_mutex_t mu = PTHREAD_MUTEX_INITIALIZER;
pthread_t thread1, thread2;
void* thread_1_proc(void*); //thread1's startup routine, tank up the bucket
void* thread_2_proc(void*); //thread2's startup routine, read the bucket
int main()
{
pthread_create(&thread1, NULL, thread_1_proc, NULL);
pthread_create(&thread2, NULL, thread_2_proc, NULL);
pthread_join(thread1);
pthread_join(thread2);
}
Below is my implementation for thread_x_proc:
void* thread_1_proc(void*)
{
while(1) { //make it work forever
pthread_mutex_lock(&mu); //lock the mutex, right?
cout << "tanking\n";
for(int i=0; i<5; i++)
bucket[i] = rand(); //actually, rand() returns int, doesn't matter
pthread_mutex_unlock(&mu); //bucket tanked, unlock the mutex, right?
//sleep(1); /* this line is commented */
}
}
void* thread_2_proc(void*)
{
while(1) {
pthread_mutex_lock(&mu);
cout << "reading\n";
for(int i=0; i<5; i++)
cout << bucket[i] << " "; //read each element in the bucket
pthread_mutex_unlock(&mu); //reading done, unlock the mutex, right?
//sleep(1); /* this line is commented */
}
}
Question
Is my implementation right? Cuz the output is not as what I expected.
...
reading
5.09434e+08 6.58441e+08 1.2288e+08 8.16198e+07 4.66482e+07 7.08736e+08 1.33455e+09
reading
5.09434e+08 6.58441e+08 1.2288e+08 8.16198e+07 4.66482e+07 7.08736e+08 1.33455e+09
reading
5.09434e+08 6.58441e+08 1.2288e+08 8.16198e+07 4.66482e+07 7.08736e+08 1.33455e+09
reading
tanking
tanking
tanking
tanking
...
But if I uncomment the sleep(1); in each thread_x_proc function, the output is right, tanking and reading follow each other, like this:
...
tanking
reading
1.80429e+09 8.46931e+08 1.68169e+09 1.71464e+09 1.95775e+09 4.24238e+08 7.19885e+08
tanking
reading
1.64976e+09 5.96517e+08 1.18964e+09 1.0252e+09 1.35049e+09 7.83369e+08 1.10252e+09
tanking
reading
2.0449e+09 1.96751e+09 1.36518e+09 1.54038e+09 3.04089e+08 1.30346e+09 3.50052e+07
...
Why? Should I use sleep() when using mutex?
Your code is technically correct, but it does not make a lot of sense, and it does not do what you assume.
What your code does is, it updates a section of data atomically, and reads from that section, atomically. However, you don't know in which order this happens, nor how often the data is written to before being read (or if at all!).
What you probably wanted is generate exactly one sequence of numbers in one thread every time and read exactly one new sequence each time in the other thread. For this, you would use either have to use an additional semaphore or better a single-producer-single-consumer queue.
In general the answer to "when should I use a mutex" is "never, if you can help it". Threads should send messages, not share state. This makes a mutex most of the time unnecessary, and offers parallelism (which is the main incentive for using threads in the first place).
The mutex makes your threads run lockstep, so you could as well just run in a single thread.
There is no implied order in which threads will get to run. This means you shall not expect any order. What's more it is possible to get on thread running over and over without letting the other to run. This is implementation specific and should be assumed random.
The case you presented falls much rather for a semaphor which is "posted" with each element added.
However if it has always to be like:
write 5 elements
read 5 elements
you should have two mutexes:
one that blocks producer until the consumer finished
one that blocks consumer until the producer finished
So the code should look something like that:
Producer:
while(true){
lock( &write_mutex )
[insert data]
unlock( &read_mutex )
}
Consumer:
while(true){
lock( &read_mutex )
[insert data]
unlock( &write_mutex )
}
Initially write_mutex should be unlocked and read_mutex locked.
As I said your code seems to be a better case for semaphores or maybe condition variables.
Mutexes are not meant for cases such as this (which doesn't mean you can't use them, it just means there are more handy tools to solve that problem).
You have no right to assume that just because you want your threads to run in a particular order, the implementation will figure out what you want and actually run them in that order.
Why shouldn't thread2 run before thread1? And why shouldn't each thread complete its loop several times before the other thread gets a chance to run up to the line where it acquires the mutex?
If you want execution to switch between two threads in a predictable way, then you need to use a semaphore, condition variable, or other mechanism for messaging between the two threads. sleep appears to result in the order you want on this occasion, but even with the sleep you haven't done enough to guarantee that they will alternate. And I have no idea why the sleep makes a difference to which thread gets to run first -- is that consistent across several runs?
If you have two functions that should execute sequentially, i.e. F1 should finish before F2 starts, then you shouldn't be using two threads. Run F2 on the same thread as F1, after F1 returns.
Without threads, you won't need the mutex either.
It isn't really the issue here.
The sleep only lets the 'other' thread access the mutex lock (by chance, it is waiting for the lock so Probably it will have the mutex), there is no way you can be sure the first thread won't re-lock the mutex though and let the other thread access it.
Mutex is for protecting data so two threads don't :
a) write simultaneously
b) one is writing when another is reading
It is not for making threads work in a certain order (if you want that functionality, ditch the threaded approach or use a flag to tell that the 'tank' is full for example).
By now, it should be clear, from the other answers, what are the mistakes in the original code. So, let's try to improve it:
/* A flag that indicates whose turn it is. */
char tanked = 0;
void* thread_1_proc(void*)
{
while(1) { //make it work forever
pthread_mutex_lock(&mu); //lock the mutex
if(!tanked) { // is it my turn?
cout << "tanking\n";
for(int i=0; i<5; i++)
bucket[i] = rand(); //actually, rand() returns int, doesn't matter
tanked = 1;
}
pthread_mutex_unlock(&mu); // unlock the mutex
}
}
void* thread_2_proc(void*)
{
while(1) {
pthread_mutex_lock(&mu);
if(tanked) { // is it my turn?
cout << "reading\n";
for(int i=0; i<5; i++)
cout << bucket[i] << " "; //read each element in the bucket
tanked = 0;
}
pthread_mutex_unlock(&mu); // unlock the mutex
}
}
The code above should work as expected. However, as others have pointed out, the result would be better accomplished with one of these two other options:
Sequentially. Since the producer and the consumer must alternate, you don't need two threads. One loop that tanks and then reads would be enough. This solution would also avoid the busy waiting that happens in the code above.
Using semaphores. This would be the solution if the producer was able to run several times in a row, accumulating elements in a bucket (not the case in the original code, though).
http://en.wikipedia.org/wiki/Producer-consumer_problem#Using_semaphores