I'm doing a C application that reads and parses data from a set of sensors and, according to the readings of the senors, it turns on or off actuators.
For my application I will be using two threads, one to read and parse the data from the sensors and another one to act on the actuators. Obviously we may face the problem of one thread reading data from a certain variable while another one is trying to write on it. This is a sample code.
#include <pthread.h>
int sensor_values;
void* reads_from_sensor(){
//writes on sensor_values, while(1) loop
}
void* turns_on_or_off(){
//reads from sensor_values, while(1) loop
}
int main(){
pthread_t threads[2];
pthread_create(&threads[1],NULL,reads_from_sensor,NULL);
pthread_create(&threads[2],NULL,turns_on_or_off,NULL);
//code continues after
}
My question is how I can solve this issue, of a certain thread writing on a certain global variable while other thread is trying to read from it, at the same time. Thanks in advance.
OP wrote in the comments
The project is still in an alpha stage. I'll make sure I optimize it once it is done. #Pablo, the shared variable is sensor_values. reads_from_sensors write on it and turns_on_or_off reads from it.
...
sensor_value would be a float as it stores a value measured by a certain sensor. That value can either be voltage, temperature or humidity
In that case I'd use conditional variables using pthread_cond_wait and
pthread_cond_signal. With these functions you can synchronize threads
with each other.
The idea is that both threads get a pointer to a mutx, the condition variable
and the shared resource, whether you declared them a global or you pass them as
thread arguments, doesn't change the idea. In the code below I'm passing all
of these as thread arguments, because I don't like global variables.
The reading thread would lock the mutex and when it reads a new value of the
sensor, it writes the new value in the shared resource. Then it call
pthread_cond_signal to send a signal to the turning thread that a new value
arrived and that it can read from it.
The turning thread would also lock the mutex and execute pthread_cond_wait to
wait on the signal. The locking must be done in that way, because
pthread_cond_wait will release the lock and make the thread block until the
signal is sent:
man pthread_cond_wait
DESCRIPTION
The pthread_cond_timedwait() and pthread_cond_wait() functions shall block on a condition variable. The application shall ensure that
these functions are called with mutex locked by the calling thread; otherwise, an error (for PTHREAD_MUTEX_ERRORCHECK and robust
mutexes) or undefined behavior (for other mutexes) results.
These functions atomically release mutex and cause the calling thread to block on the condition variable cond; atomically here means
atomically with respect to access by another thread to the mutex and then the condition variable. That is, if another thread is
able to acquire the mutex after the about-to-block thread has released it, then a subsequent call to pthread_cond_broadcast() or
pthread_cond_signal() in that thread shall behave as if it were issued after the about-to-block thread has blocked.
Example:
#include <stdio.h>
#include <pthread.h>
#include <stdlib.h>
#include <time.h>
#include <unistd.h>
struct thdata {
pthread_mutex_t *mutex;
pthread_cond_t *cond;
int *run;
float *sensor_value; // the shared resource
};
void *reads_from_sensors(void *tdata)
{
struct thdata *data = tdata;
int i = 0;
while(*data->run)
{
pthread_mutex_lock(data->mutex);
// read from sensor
*data->sensor_value = (rand() % 2000 - 1000) / 10.0;
// just for testing, send a singnal only every
// 3 reads
if((++i % 3) == 0)
{
printf("read: value == %f, sending signal\n", *data->sensor_value);
pthread_cond_signal(data->cond);
}
pthread_mutex_unlock(data->mutex);
sleep(1);
}
// sending signal so that other thread can
// exit
pthread_mutex_lock(data->mutex);
pthread_cond_signal(data->cond);
pthread_mutex_unlock(data->mutex);
puts("read: bye");
pthread_exit(NULL);
}
void *turns_on_or_off (void *tdata)
{
struct thdata *data = tdata;
while(*data->run)
{
pthread_mutex_lock(data->mutex);
pthread_cond_wait(data->cond, data->mutex);
printf("turns: value read: %f\n\n", *data->sensor_value);
pthread_mutex_unlock(data->mutex);
usleep(1000);
}
puts("turns: bye");
pthread_exit(NULL);
}
int main(void)
{
srand(time(NULL));
struct thdata thd[2];
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
// controlling vars
int run_rfs = 1;
int run_tof = 1;
float sensor_value;
thd[0].run = &run_rfs;
thd[1].run = &run_tof;
thd[0].mutex = &mutex;
thd[1].mutex = &mutex;
thd[0].cond = &cond;
thd[1].cond = &cond;
thd[0].sensor_value = &sensor_value;
thd[1].sensor_value = &sensor_value;
pthread_t th[2];
printf("Press ENTER to exit...\n");
pthread_create(th, NULL, reads_from_sensors, thd);
pthread_create(th + 1, NULL, turns_on_or_off, thd + 1);
getchar();
puts("Stopping threads...");
run_rfs = 0;
run_tof = 0;
pthread_join(th[0], NULL);
pthread_join(th[1], NULL);
return 0;
}
Output:
$ ./a
Press ENTER to exit...
read: value == -99.500000, sending signal
turns: value read: -99.500000
read: value == -25.200001, sending signal
turns: value read: -25.200001
read: value == 53.799999, sending signal
turns: value read: 53.799999
read: value == 20.400000, sending signal
turns: value read: 20.400000
Stopping threads...
read: bye
turns: value read: 20.400000
turns: bye
Note that in the example I only send the signal every 3 seconds (and do a long
sleep(1)) for testing purposes, otherwise the terminal would overflow immediately
and you would have a hard time reading the output.
See also: understanding of pthread_cond_wait() and pthread_cond_signal()
Your question is too generic. There are different multithread synchronization methods mutex, reader-writer locks, conditional variables and so on.
The easiest and most simple are mutex (mutual excluasion). They are pthread_mutex_t type variables. You first need to initialize them; you can do it in two ways:
assigning to the mutex variable the constant value PTHREAD_MUTEX_INITIALIZER
calling the funtion pthread_mutex_init
Then before reading or writing a shared variable you call the function int pthread_mutex_lock(pthread_mutex_t *mutex); and after exited the critical section you must release the critical section by calling int pthread_mutex_unlock(pthread_mutex_t *mutex);.
If the resource is busy the lock will block the execution of your code until it gets released. If you want to avoid that take a look at int pthread_mutex_trylock(pthread_mutex_t *mutex);.
If your program has much more reads than writes on the same shared variable, take a look at the Reader-Writer locks.
Related
I'm writing a program which receive data from websocket and work with this data in thread pool.
I have problem with pthread_cond_wait when processor have 2 or more cores. After pthread_cond_signal signal is received by all threads which run on different cores. For example if I have 2 cores, then the signal will come to 2 threads at once, which are located on these two cores. If I have single core processor all is good.
What I have to do to get the program to work correctly on multi-core processors? So that only one thread receives the signal to start work.
I wrote an example of my code with generation random text data instead of websocket data.
#include<stdio.h>
#include<stdlib.h>
#include<cstring>
#include<pthread.h>
#include<unistd.h>
pthread_attr_t attrd;
pthread_mutex_t mutexQueue;
pthread_cond_t condQueue;
char textArr[128][24]; //array with random text to work
int tc; //tasks count
int gi; //global array index
void *workThread(void *args){
int ai;//internal index for working array element
while(1){
pthread_mutex_lock(&mutexQueue);
while(tc==0){
pthread_cond_wait(&condQueue,&mutexQueue); //wait for signal if tasks count = 0.
}
ai=gi;
if(gi==127)gi=0;else gi++;
tc--;
pthread_mutex_unlock(&mutexQueue);
printf("%s\r\n",textArr[ai]);
// then work with websocket data
}
}
void *generalThread(void *args){
const char chrs[]="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"; //chars fo random text generation
int ai=0;
srand(time(NULL));
while(1){
for(int i=0;i<23;i++)textArr[ai][i]=chrs[rand()%61];//generating data instead of websocket data
textArr[ai][23]='\0';
tc++;
pthread_cond_signal(&condQueue); //Send signal for thread to begin work with data
if(ai==127)ai=0;else ai++;
}
}
int main(int argc,char *argv[]){
pthread_attr_init(&attrd);
pthread_attr_setdetachstate(&attrd,PTHREAD_CREATE_DETACHED);
pthread_t gt,wt[32];
for(int i=0;i<32;i++)pthread_create(&wt[i],&attrd,&workThread,NULL);
pthread_create(>,NULL,&generalThread,NULL);
pthread_join(gt,NULL);
return 0;
}
First some info:
man pthread_cond_wait
Rationale
Some implementations, particularly on a multi-processor, may sometimes cause multiple threads to wake up when the condition variable is signaled simultaneously on different processors.
man pthread_cond_signal
Rationale
Multiple Awakenings by Condition Signal
On a multi-processor, it may be impossible for an implementation of pthread_cond_signal() to avoid the unblocking of more than one thread blocked on a condition variable.
...
The effect is that more than one thread can return from its call to pthread_cond_wait() or pthread_cond_timedwait() as a result of one call to pthread_cond_signal(). This effect is called "spurious wakeup". Note that the situation is self-correcting in that the number of threads that are so awakened is finite; for example, the next thread to call pthread_cond_wait() after the sequence of events above blocks.
So far, so good, the code in your workThread is proper synchronized (but you should put the printf in the synchronized section as well) but the code in your generalThread has no synchronization at all. Encapsulate the code in the while loop with a lock / unlock.
In that case, the first awakened thread has to aquire a lock on the specified mutex, which will be owned by either another thread or the generalThread. Until the mutex is unlocked, the thread blocks (no matter the reason of its wakeup). After the aquisition, it owns the mutex and all other threads will be blocked, the generalThread inclusive.
Note: a pthread_cond_wait implicitly unlocks the specified mutex upon entering the wait state and on a wakeup it tries to aquire a lock on the specified mutex.
Adding a mutex lock to tc++ fully corrects my programs:
void *generalThread(void *args) {
const char chrs[]="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
int ai=0;
srand(time(NULL));
while(1){
for(int i=0;i<23;i++)textArr[ai][i]=chrs[rand()%61];
textArr[ai][23]='\0';
pthread_mutex_lock(&mutexQueue); //this has been added
tc++;
pthread_mutex_unlock(&mutexQueue); //this has been added
pthread_cond_signal(&condQueue);
if(ai==127)ai=0;else ai++;
}
}
I'm studying on condition variables of Pthread. When I'm reading the explanation of pthread_cond_signal, I see the following.
The pthread_cond_signal() function shall unblock at least one of
the
threads that are blocked on the specified condition variable cond (if
any threads are blocked on cond).
Till now I knew pthread_cond_signal() would make only one thread to wake up at a time. But, the quoted explanation says at least one. What does it mean? Can it make more than one thread wake up? If yes, why is there pthread_cond_broadcast()?
En passant, I wish the following code taken from UNIX Systems Programming book of Robbins is also related to my question. Is there any reason the author's pthread_cond_broadcast() usage instead of pthread_cond_signal() in waitbarrier function? As a minor point, why is !berror checking needed too as a part of the predicate? When I try both of them by changing, I cannot see any difference.
/*
The program implements a thread-safe barrier by using condition variables. The limit
variable specifies how many threads must arrive at the barrier (execute the
waitbarrier) before the threads are released from the barrier.
The count variable specifies how many threads are currently waiting at the barrier.
Both variables are declared with the static attribute to force access through
initbarrier and waitbarrier. If successful, the initbarrier and waitbarrier
functions return 0. If unsuccessful, these functions return a nonzero error code.
*/
#include <errno.h>
#include <pthread.h>
#include <stdio.h>
static pthread_cond_t bcond = PTHREAD_COND_INITIALIZER;
static pthread_mutex_t bmutex = PTHREAD_MUTEX_INITIALIZER;
static int count = 0;
static int limit = 0;
int initbarrier(int n) { /* initialize the barrier to be size n */
int error;
if (error = pthread_mutex_lock(&bmutex)) /* couldn't lock, give up */
return error;
if (limit != 0) { /* barrier can only be initialized once */
pthread_mutex_unlock(&bmutex);
return EINVAL;
}
limit = n;
return pthread_mutex_unlock(&bmutex);
}
int waitbarrier(void) { /* wait at the barrier until all n threads arrive */
int berror = 0;
int error;
if (error = pthread_mutex_lock(&bmutex)) /* couldn't lock, give up */
return error;
if (limit <= 0) { /* make sure barrier initialized */
pthread_mutex_unlock(&bmutex);
return EINVAL;
}
count++;
while ((count < limit) && !berror)
berror = pthread_cond_wait(&bcond, &bmutex);
if (!berror) {
fprintf(stderr,"soner %d\n",
(int)pthread_self());
berror = pthread_cond_broadcast(&bcond); /* wake up everyone */
}
error = pthread_mutex_unlock(&bmutex);
if (berror)
return berror;
return error;
}
/* ARGSUSED */
static void *printthread(void *arg) {
fprintf(stderr,"This is the first print of thread %d\n",
(int)pthread_self());
waitbarrier();
fprintf(stderr,"This is the second print of thread %d\n",
(int)pthread_self());
return NULL;
}
int main(void) {
pthread_t t0,t1,t2;
if (initbarrier(3)) {
fprintf(stderr,"Error initilizing barrier\n");
return 1;
}
if (pthread_create(&t0,NULL,printthread,NULL))
fprintf(stderr,"Error creating thread 0.\n");
if (pthread_create(&t1,NULL,printthread,NULL))
fprintf(stderr,"Error creating thread 1.\n");
if (pthread_create(&t2,NULL,printthread,NULL))
fprintf(stderr,"Error creating thread 2.\n");
if (pthread_join(t0,NULL))
fprintf(stderr,"Error joining thread 0.\n");
if (pthread_join(t1,NULL))
fprintf(stderr,"Error joining thread 1.\n");
if (pthread_join(t2,NULL))
fprintf(stderr,"Error joining thread 2.\n");
fprintf(stderr,"All threads complete.\n");
return 0;
}
Due to spurious wake-ups pthread_cond_signal could wake up more than one thread.
Look for word "spurious" in pthread_cond_wait.c from glibc.
In waitbarrier it must wake up all threads when they all have arrived to that point, hence it uses pthread_cond_broadcast.
Can [pthread_cond_signal()] make more than one thread wake up?
That's not guaranteed. On some operating system, on some hardware platform, under some circumstances it could wake more than one thread. It is allowed to wake more than one thread because that gives the implementer more freedom to make it work in the most efficient way possible for any given hardware and OS.
It must wake at least one waiting thread, because otherwise, what would be the point of calling it?
But, if your applicaton needs a signal that is guaranteed to wake all of the waiting threads, then that is what pthread_cond_broadcast() is for.
Making efficient use of a multi-processor system is hard. https://www.e-reading.club/bookreader.php/134637/Herlihy,Shavit-_The_art_of_multiprocessor_programming.pdf
Most programming language and library standards allow similar freedoms in the behavior of multi-threaded programs, for the same reason: To allow programs to achieve high performance on a variety of different platforms.
I am beginner to SO, so please let me know if the question is not clear.
I am using two threads for example A and B. And i have a global variable 'p'.
Thread A is while looping and incrementing the value of 'p'.At the same time B is trying to set the 'p' with some other value(both are two different thread functions).
If I am using mutex for synchronizations, and the thread A get the mutex and incrementation the 'p' in a while loop,but it does not release the mutex.
So my question is that if the thread A doesn’t release the mutex can the thread B access the variable 'p'??
EDIT
The thread B is also protected accses to 'p' using mutex.
If the thread A lock using pthread_mutex_lock(), and doesn’t release it , then what happen if the same thread(A) try to access the lock again(remember the thread A is while looping)
For example
while(1)
{
pthread_mutex_lock(&mutex);
p = 10;
}
Is there any problem with this code if the mutex is never released?
You can still access the variable in thread B as the mutex is a separate object not connected to the variable. If You call mutex lock from thread B before accessing p then the thread B will wait for mutex to be released. In fact the thread A will only execute loop body once as it will wait for the mutex to be released before it can lock it again.
If You don't unlock the mutex then any call to lock the same mutex will wait indefinitely, but the variable will be writable.
In Your example access to variable p is what is called a critical section, or the part of code that is between mutex lock and mutex release.
There is no restriction on mutex, you need to write your program to following the rules of using mutex.
Here is the basic steps to use mutex on shared resource:
Acquire lock first
do job (increase for A, set value for B)
Release lock,
If both A & B follow the rules, then B can't modify it, while A keeps the lock.
Or, if your thread B don't acquire the lock first, it of cause could modify the variable, but that would be a bug for concurrent programming.
And, by the way, you can also use condition together with mutex, so that you can let threads wait & notify each other, instead of looping all the time which is a waste of machine resource.
For your updated question
On linux, in c, there are mainly 3 methods to acquire lock of mutex, what happens when a thread can't get the lock depends on which methods u use.
int pthread_mutex_lock(pthread_mutex_t * mutex );
if it's already locked by another thread, then it block until the lock is unlocked,
int pthread_mutex_trylock(pthread_mutex_t * mutex );
similar to pthread_mutex_lock(), but it won't block, instead return error EBUSY,
int pthread_mutex_timedlock(pthread_mutex_t *restrict mutex, const struct timespec *restrict abs_timeout);
similar to pthread_mutex_lock(), but it will wait for a timeout before return error ETIMEDOUT,
Simple example of statically initialized mutex
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
static int p = 0;
static pthread_mutex_t locker = PTHREAD_MUTEX_INITIALIZER;
static void *
threadFunc(void *arg)
{
int err;
err = pthread_mutex_lock(&locker);
if (err != 0){
perror("pthread_mutex_lock failed");
exit(1);
}
p++;
err = pthread_mutex_unlock(&locker);
if (err != 0){
perror("pthread_mutex_unlock failed");
exit(1);
}
return NULL;
}
int
main(int argc, char *argv[])
{
pthread_t A, B;
pthread_create(&A, NULL, threadFunc, NULL);
pthread_create(&B, NULL, threadFunc, NULL);
pthread_join(A, NULL);
pthread_join(B, NULL);
printf("p = %d\n", p);
return 0;
}
Error checking in main is omitted for brevity but should be used. If you do not release mutex program will never finish, thread B will never get lock.
I am trying to write a code that does not block main() when pthread_join() is called:
i.e. basically trying to implement my previous question mentioned below:
https://stackoverflow.com/questions/24509500/pthread-join-and-main-blocking-multithreading
And the corresponding explanation at:
pthreads - Join on group of threads, wait for one to exit
As per suggested answer:
You'd need to create your own version of it - e.g. an array of flags (one flag per thread) protected by a mutex and a condition variable; where just before "pthread_exit()" each thread acquires the mutex, sets its flag, then does "pthread_cond_signal()". The main thread waits for the signal, then checks the array of flags to determine which thread/s to join (there may be more than one thread to join by then).
I have tried as below:
My status array which keeps a track of which threads have finished:
typedef struct {
int Finish_Status[THREAD_NUM];
int signalled;
pthread_mutex_t mutex;
pthread_cond_t FINISHED;
}THREAD_FINISH_STATE;
The thread routine, it sets the corresponding array element when the thread finishes and also signals the condition variable:
void* THREAD_ROUTINE(void* arg)
{
THREAD_ARGUMENT* temp=(THREAD_ARGUMENT*) arg;
printf("Thread created with id %d\n",temp->id);
waitFor(5);
pthread_mutex_lock(&(ThreadFinishStatus.mutex));
ThreadFinishStatus.Finish_Status[temp->id]=TRUE;
ThreadFinishStatus.signalled=TRUE;
if(ThreadFinishStatus.signalled==TRUE)
{
pthread_cond_signal(&(ThreadFinishStatus.FINISHED));
printf("Signal that thread %d finished\n",temp->id);
}
pthread_mutex_unlock(&(ThreadFinishStatus.mutex));
pthread_exit((void*)(temp->id));
}
I am not able to write the corresponding parts pthread_join() and pthread_cond_wait() functions. There are a few things which I am not able to implement.
1) How to write corresponding part pthread_cond_wait() in my main()?
2) I am trying to write it as:
pthread_mutex_lock(&(ThreadFinishStatus.mutex));
while((ThreadFinishStatus.signalled != TRUE){
pthread_cond_wait(&(ThreadFinishStatus.FINISHED), &(ThreadFinishStatus.mutex));
printf("Main Thread signalled\n");
ThreadFinishStatus.signalled==FALSE; //Reset signalled
//check which thread to join
}
pthread_mutex_unlock(&(ThreadFinishStatus.mutex));
But it does not enter the while loop.
3) How to use pthread_join() so that I can get the return value stored in my arg[i].returnStatus
i.e. where to put below statement in my main:
`pthread_join(T[i],&(arg[i].returnStatus));`
COMPLETE CODE
#include <stdio.h>
#include <pthread.h>
#include <time.h>
#define THREAD_NUM 5
#define FALSE 0
#define TRUE 1
void waitFor (unsigned int secs) {
time_t retTime;
retTime = time(0) + secs; // Get finishing time.
while (time(0) < retTime); // Loop until it arrives.
}
typedef struct {
int Finish_Status[THREAD_NUM];
int signalled;
pthread_mutex_t mutex;
pthread_cond_t FINISHED;
}THREAD_FINISH_STATE;
typedef struct {
int id;
void* returnStatus;
}THREAD_ARGUMENT;
THREAD_FINISH_STATE ThreadFinishStatus;
void initializeState(THREAD_FINISH_STATE* state)
{
int i=0;
state->signalled=FALSE;
for(i=0;i<THREAD_NUM;i++)
{
state->Finish_Status[i]=FALSE;
}
pthread_mutex_init(&(state->mutex),NULL);
pthread_cond_init(&(state->FINISHED),NULL);
}
void destroyState(THREAD_FINISH_STATE* state)
{
int i=0;
for(i=0;i<THREAD_NUM;i++)
{
state->Finish_Status[i]=FALSE;
}
pthread_mutex_destroy(&(state->mutex));
pthread_cond_destroy(&(state->FINISHED));
}
void* THREAD_ROUTINE(void* arg)
{
THREAD_ARGUMENT* temp=(THREAD_ARGUMENT*) arg;
printf("Thread created with id %d\n",temp->id);
waitFor(5);
pthread_mutex_lock(&(ThreadFinishStatus.mutex));
ThreadFinishStatus.Finish_Status[temp->id]=TRUE;
ThreadFinishStatus.signalled=TRUE;
if(ThreadFinishStatus.signalled==TRUE)
{
pthread_cond_signal(&(ThreadFinishStatus.FINISHED));
printf("Signal that thread %d finished\n",temp->id);
}
pthread_mutex_unlock(&(ThreadFinishStatus.mutex));
pthread_exit((void*)(temp->id));
}
int main()
{
THREAD_ARGUMENT arg[THREAD_NUM];
pthread_t T[THREAD_NUM];
int i=0;
initializeState(&ThreadFinishStatus);
for(i=0;i<THREAD_NUM;i++)
{
arg[i].id=i;
}
for(i=0;i<THREAD_NUM;i++)
{
pthread_create(&T[i],NULL,THREAD_ROUTINE,(void*)&arg[i]);
}
/*
Join only if signal received
*/
pthread_mutex_lock(&(ThreadFinishStatus.mutex));
//Wait
while((ThreadFinishStatus.signalled != TRUE){
pthread_cond_wait(&(ThreadFinishStatus.FINISHED), &(ThreadFinishStatus.mutex));
printf("Main Thread signalled\n");
ThreadFinishStatus.signalled==FALSE; //Reset signalled
//check which thread to join
}
pthread_mutex_unlock(&(ThreadFinishStatus.mutex));
destroyState(&ThreadFinishStatus);
return 0;
}
Here is an example of a program that uses a counting semaphore to watch as threads finish, find out which thread it was, and review some result data from that thread. This program is efficient with locks - waiters are not spuriously woken up (notice how the threads only post to the semaphore after they've released the mutex protecting shared state).
This design allows the main program to process the result from some thread's computation immediately after the thread completes, and does not require the main wait for all threads to complete. This would be especially helpful if the running time of each thread varied by a significant amount.
Most importantly, this program does not deadlock nor race.
#include <pthread.h>
#include <semaphore.h>
#include <stdlib.h>
#include <stdio.h>
#include <queue>
void* ThreadEntry(void* args );
typedef struct {
int threadId;
pthread_t thread;
int threadResult;
} ThreadState;
sem_t completionSema;
pthread_mutex_t resultMutex;
std::queue<int> threadCompletions;
ThreadState* threadInfos;
int main() {
int numThreads = 10;
int* threadResults;
void* threadResult;
int doneThreadId;
sem_init( &completionSema, 0, 0 );
pthread_mutex_init( &resultMutex, 0 );
threadInfos = new ThreadState[numThreads];
for ( int i = 0; i < numThreads; i++ ) {
threadInfos[i].threadId = i;
pthread_create( &threadInfos[i].thread, NULL, &ThreadEntry, &threadInfos[i].threadId );
}
for ( int i = 0; i < numThreads; i++ ) {
// Wait for any one thread to complete; ie, wait for someone
// to queue to the threadCompletions queue.
sem_wait( &completionSema );
// Find out what was queued; queue is accessed from multiple threads,
// so protect with a vanilla mutex.
pthread_mutex_lock(&resultMutex);
doneThreadId = threadCompletions.front();
threadCompletions.pop();
pthread_mutex_unlock(&resultMutex);
// Announce which thread ID we saw finish
printf(
"Main saw TID %d finish\n\tThe thread's result was %d\n",
doneThreadId,
threadInfos[doneThreadId].threadResult
);
// pthread_join to clean up the thread.
pthread_join( threadInfos[doneThreadId].thread, &threadResult );
}
delete threadInfos;
pthread_mutex_destroy( &resultMutex );
sem_destroy( &completionSema );
}
void* ThreadEntry(void* args ) {
int threadId = *((int*)args);
printf("hello from thread %d\n", threadId );
// This can safely be accessed since each thread has its own space
// and array derefs are thread safe.
threadInfos[threadId].threadResult = rand() % 1000;
pthread_mutex_lock( &resultMutex );
threadCompletions.push( threadId );
pthread_mutex_unlock( &resultMutex );
sem_post( &completionSema );
return 0;
}
Pthread conditions don't have "memory"; pthread_cond_wait doesn't return if pthread_cond_signal is called before pthread_cond_wait, which is why it's important to check the predicate before calling pthread_cond_wait, and not call it if it's true. But that means the action, in this case "check which thread to join" should only depend on the predicate, not on whether pthread_cond_wait is called.
Also, you might want to make the while loop actually wait for all the threads to terminate, which you aren't doing now.
(Also, I think the other answer about "signalled==FALSE" being harmless is wrong, it's not harmless, because there's a pthread_cond_wait, and when that returns, signalled would have changed to true.)
So if I wanted to write a program that waited for all threads to terminate this way, it would look more like
pthread_mutex_lock(&(ThreadFinishStatus.mutex));
// AllThreadsFinished would check that all of Finish_Status[] is true
// or something, or simpler, count the number of joins completed
while (!AllThreadsFinished()) {
// Wait, keeping in mind that the condition might already have been
// signalled, in which case it's too late to call pthread_cond_wait,
// but also keeping in mind that pthread_cond_wait can return spuriously,
// thus using a while loop
while (!ThreadFinishStatus.signalled) {
pthread_cond_wait(&(ThreadFinishStatus.FINISHED), &(ThreadFinishStatus.mutex));
}
printf("Main Thread signalled\n");
ThreadFinishStatus.signalled=FALSE; //Reset signalled
//check which thread to join
}
pthread_mutex_unlock(&(ThreadFinishStatus.mutex));
Your code is racy.
Suppose you start a thread and it finishes before you grab the mutex in main(). Your while loop will never run because signalled was already set to TRUE by the exiting thread.
I will echo #antiduh's suggestion to use a semaphore that counts the number of dead-but-not-joined threads. You then loop up to the number of threads spawned waiting on the semaphore. I'd point out that the POSIX sem_t is not like a pthread_mutex in that sem_wait can return EINTR.
Your code appears fine. You have one minor buglet:
ThreadFinishStatus.signalled==FALSE; //Reset signalled
This does nothing. It tests whether signalled is FALSE and throws away the result. That's harmless though since there's nothing you need to do. (You never want to set signalled to FALSE because that loses the fact that it was signalled. There is never any reason to unsignal it -- if a thread finished, then it's finished forever.)
Not entering the while loop means signalled is TRUE. That means the thread already set it, in which case there is no need to enter the loop because there's nothing to wait for. So that's fine.
Also:
ThreadFinishStatus.signalled=TRUE;
if(ThreadFinishStatus.signalled==TRUE)
There's no need to test the thing you just set. It's not like the set can fail.
FWIW, I would suggest re-architecting. If the existing functions like pthread_join don't do exactly what you want, just don't use them. If you're going to have structures that track what work is done, then totally separate that from thread termination. Since you will already know what work is done, what different does it make when and how threads terminate? Don't think of this as "I need a special way to know when a thread terminates" and instead think of this "I need to know what work is done so I can do other things".
Writing my basic programs on multi threading and I m coming across several difficulties.
In the program below if I give sleep at position 1 then value of shared data being printed is always 10 while keeping sleep at position 2 the value of shared data is always 0.
Why this kind of output is coming ?
How to decide at which place we should give sleep.
Does this mean that if we are placing a sleep inside the mutex then the other thread is not being executed at all thus the shared data being 0.
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include<unistd.h>
pthread_mutex_t lock;
int shared_data = 0;
void * function(void *arg)
{
int i ;
for(i =0; i < 10; i++)
{
pthread_mutex_lock(&lock);
shared_data++;
pthread_mutex_unlock(&lock);
}
pthread_exit(NULL);
}
int main()
{
pthread_t thread;
void * exit_status;
int i;
pthread_mutex_init(&lock, NULL);
i = pthread_create(&thread, NULL, function, NULL);
for(i =0; i < 10; i++)
{
sleep(1); //POSITION 1
pthread_mutex_lock(&lock);
//sleep(1); //POSITION 2
printf("Shared data value is %d\n", shared_data);
pthread_mutex_unlock(&lock);
}
pthread_join(thread, &exit_status);
pthread_mutex_destroy(&lock);
}
When you sleep before you lock the mutex, then you're giving the other thread plenty of time to change the value of the shared variable. That's why you're seeing a value of "10" with the 'sleep' in position #1.
When you grab the mutex first, you're able to lock it fast enough that you can print out the value before the other thread has a chance to modify it. The other thread sits and blocks on the pthread_mutex_lock() call until your main thread has finished sleeping and unlocked it. At that point, the second thread finally gets to run and alter the value. That's why you're seeing a value of "0" with the 'sleep' at position #2.
This is a classic case of a race condition. On a different machine, the same code might not display "0" with the sleep call at position #2. It's entirely possible that the second thread has the opportunity to alter the value of the variable once or twice before your main thread locks the mutex. A mutex can ensure that two threads don't access the same variable at the same time, but it doesn't have any control over the order in which the two threads access it.
I had a full explanation here but ended up deleting it. This is a basic synchronization problem and you should be able to trace and identify it before tackling anything more complicated.
But I'll give you a hint: It's only the sleep() in position 1 that matters; the other one inside the lock is irrelevant as long as it doesn't change the code outside the lock.