I have the following code below. I want only half of the threads to enter the threadedfunction at a time. How do I create a Semaphore to block the other processes? And how would I go about unblocking the previously blocked processes whenever the threads have finished using the function?
#include <iostream>
#include <unistd.h>
#include <sys/wait.h>
#include <pthread.h>
using namespace std;
#define NUM_THREADS 4
long int sharedcount;
pthread_mutex_t count_mutex;
//Function that will be run by multiple threads
//Needs to return a void pointer and if it takes arguments
//it needs to be a void pointer
void *ThreadedFunction(void *threadid)
{
int success;
long id = (long)threadid;
//Lock mutex preventing the other threads from ru nning
success = pthread_mutex_lock( &count_mutex );
cout << "Thread " << id << " beginning.\n";
for(int i = 0; i < 100000000; i++)
sharedcount++;
cout << "Thread " << id << " exiting.\n";
cout << sharedcount << endl;
//Unlock the mutex after the thread has finished running
pthread_mutex_unlock( &count_mutex );
//Kill the thread
pthread_exit(NULL);
}
int main ()
{
//Initialize mutex
pthread_mutex_init(&count_mutex, NULL);
//Create an array of threads
pthread_t threads[NUM_THREADS];
int rc;
int i;
sharedcount = 0;
for( i=0; i < NUM_THREADS; i++ )
{
cout << "main() : creating thread, " << i << endl;
//Create thread by storing it in a location in the array. Call the
//function for the threads to run inside. And pass the argument (if any).
//If no arguments pass NULL
rc = pthread_create(&threads[i], NULL, ThreadedFunction, (void *)i);
if (rc)
{
cout << "Error:unable to create thread," << rc << endl;
exit(-1);
}
}
//Have main thread wait for all other threads to stop running.
for(i = 0; i < NUM_THREADS; i++)
pthread_join(threads[i], NULL);
//cout << sharedcount << endl;
pthread_exit(NULL);
}
What you could do is use a counting semaphore (as opposed to a binary semaphore). A counting semaphore has an initial value greater than 1, allowing for multiple threads to call "wait" on the semaphore and not have those threads actually blocked and put in the semaphore queue.
What I would have done in your case is initialize a semaphore in the main function with an initial value of NUM_THREADS/2. Then I would insert a line at the beginning of threadedFunction where I do a wait(semaphore) and a line at the end of the function where you do a signal(semaphore). This way, when a thread is about to exit the function, it signals a thread that was blocked after having called wait on the semaphore and it lets this thread in.
Hope this helps.
Related
I want a thread to run infinity times in order to execute the task described on do_work() function that it receives. However, the function is only called on the pthread_create() subroutine.
I've tried to implement the sched_yield() and the pthread_join() routines on a while loop. But it didn't work yet.
Is there any routine in which I can call the existing thread again?
int main (int argc, char ** argv) {
int period;
int priority;
int load;
char schedule[15];
period = atoi(argv[1]);
priority = atoi(argv[2]);
load = atoi(argv[3]);
strncpy(schedule,argv[4],100);
std::cout << " period : " << period <<"\n priority : "<< priority << "\n load : "<< load << "\n schedule : " << schedule <<std::endl;
struct sched_param param;
pthread_t thread;
int rc;
sched_setscheduler (0, SCHED_FIFO , ¶m);
std::cout << "main() : creating thread " << std::endl;
rc = pthread_create(&thread, NULL, do_work, (void*)load);
if (rc) {
std::cout << "Error:unable to create thread " << rc << std::endl;
exit(-1);
}
int i=0;
struct sigaction action;
struct itimerval timer;
while(i<10000){
pthread_join(thread, NULL);
sched_yield();
i++;
}
pthread_exit(NULL);
}
You do not call a thread, you create a thread. By doing that, you specify a start_routine which will be called 'in' the new thread.
If you want to call repeatedly a function in a loop, then you can do the following in your start_routine:
void* start_routine(void *arg) {
while (active) { // active: atomic global boolean value
do_work();
}
// if no longer active,
// there could be an option to wait to become active again,
// or exit the thread
pthread_exit(NULL);
}
pthread_join() is only called, if you want to join a thread with other thread(s). pthread_join() waits until the target thread has terminated. By joining the thread, all resources are given back to the system (cleanup).
Just wanted to show you the code I implemented.
I was not sure about the meaning of Threads when I made this question and you helped me to understand that I cannot access the function on a thread multiple times, but I have to create it on each use.
My main objective was to find a way of calling the do_work() function on the reception of the signal SIGALRM. Thus, I just assumed the do_wordk() to be my Thread and used a sigaction struct to control the arrival of the signal.
If you guys want to test the code, it returns the execution time of the do_work() function and a message if the deadline set on your period was lost. The purpose of this work was to make an analogy with periodic threads.
To compile:
g++ teste.cpp -o exe -lrt
To run:
sudo taskset -c 0 ./exe 300 1 100000 F
sudo taskset -c 0 ./exe Period Priority Work_Load Scheduller_Policy
#include<signal.h>
#include<stdio.h>
#include<unistd.h>
#include<errno.h>
#include<sys/time.h>
#include<iostream>
#include<string>
#include<string.h>
long load=1;
void deadline();
void do_work();
void wakeup(int j){
struct itimerval aux;
int t1, t2;
getitimer( ITIMER_REAL, &aux); //Get initial resume time
t1 = aux.it_value.tv_usec;
//std::cout << "Hello World! Thread working |Resume Time : " <<t1<< std::endl;
do_work();
getitimer( ITIMER_REAL, &aux);
t2 = aux.it_value.tv_usec; //Get Final resume time
std::cout << "Execution time (usec): " <<t1 - t2<< std::endl;
if (t2==0){
deadline();
}
return;
}
void do_work(){
for ( int i = 0; i < load * 1000; i++) {
/* do nothing , keep counting */
}
}
void deadline() {
std::cout << "Lost deadline!" << std::endl;
}
int main (int argc, char ** argv) {
int i;
int period;
int priority;
char scheduler[5];
period = atoi(argv[1])*1000;
priority = atoi(argv[2]);
load = atoi(argv[3]);
strcpy(scheduler, argv[4]);
std::cout << " period : " << period <<"\n priority : "<< priority << "\n load : "<< load << "\n scheduler : " << scheduler <<std::endl;
struct sched_param param;
param.sched_priority = priority;
if (scheduler[0]=='F'){
int r = sched_setscheduler (0, SCHED_FIFO , ¶m);
if(r==-1){ perror("scheduller"); return 1;}
std::cout <<"FIFO scheduller: "<<r<<std::endl;
}else{
int r = sched_setscheduler (0, SCHED_RR , ¶m);
if(r==-1){ perror("scheduller"); return 1;}
std::cout <<"RR scheduller: "<<r<<std::endl;
}
struct itimerval val;
struct sigaction action;
sigset_t mask;
sigemptyset(&action.sa_mask);
action.sa_handler = wakeup;
action.sa_flags=SA_RESTART;
if(sigaction(SIGALRM, &action, 0)==-1){
perror("sigaction");
return 1;
}
val.it_interval.tv_sec=0;
val.it_interval.tv_usec=period;
val.it_value.tv_sec=0;
val.it_value.tv_usec=period;
if(setitimer(ITIMER_REAL, &val, 0)==-1){
perror("setitimer");
return 1;
}
if(sigwait( &mask, &i)==-1){
perror("sigwait");
}
return 0;
}
Finally, I am really grateful for your patience in understanding my problem. This is my first question on this community and I hope I'll improve them over time. Thank you all for your answers and the effort on helping me.
I have a set of consumers generating data, each with their own vector. A consumer, in this MWE the main function, should wait for one of the producers to create something, and then once it wakes up process all queues and then wait again.
With a mutex, the critical section is easy to create, but how to wait for any one of the producers to signal that data is available, and wake up main. The producers do not need to block each other, they could each add to their own queue, but each producer must block the consumer, and the consumer must block all the producers.
Here is my MWE, which is only missing the equivalent of Java obj.wait() and object.notify().
#include <thread>
#include <mutex>
#include <iostream>
#include <vector>
#include <unistd.h>
using namespace std;
mutex m;
vector<int> event1;
int count1 = 0;
void producer1() {
while(true) {
m.lock(); // enter critical section
cout << "producer1: " << count1 << '\n';
event1.push_back(count1++);
// TODO: wakeup main consumer loop
// in Java this would be obj.notify();
m.unlock();
sleep(1);
}
}
vector<int> event2;
int count2 = 0;
void producer2() {
while(true) {
m.lock(); // enter critical section
cout << "producer2: " << count2 << '\n';
event2.push_back(count2++);
// TODO: wakeup main consumer loop
// in Java this would be obj.notify();
m.unlock(); // enter critical section
sleep(2);
}
}
int main() {
thread t1(producer1);
thread t2(producer2);
while (true) {
// TODO: obj.wait();
// inside the critical section, get all the data from each producer
m.lock();
for (int i = 0; i < event1.size(); i++)
cout << "consumer1: " << event1[i] << '\n';
for (int i = 0; i < event2.size(); i++)
cout << "consumer2: " << event2[i] << '\n';
event1.clear();
event2.clear();
m.unlock();
sleep(10);
}
}
I have used mutex in inherited classes but seems it does not work as I expected with threads. Please have a look at below code:
#include <iostream>
#include <cstdlib>
#include <pthread.h>
// mutex::lock/unlock
#include <iostream> // std::cout
#include <thread> // std::thread
#include <chrono> // std::thread
#include <mutex> // std::mutex
typedef unsigned int UINT32t;
typedef int INT32t;
using namespace std;
class Abstract {
protected:
std::mutex mtx;
};
class Derived: public Abstract
{
public:
void* write( void* result)
{
UINT32t error[1];
UINT32t data = 34;
INT32t length = 0;
static INT32t counter = 0;
cout << "\t before Locking ..." << " in thread" << endl;
mtx.lock();
//critical section
cout << "\t After Create " << ++ counter << " device in thread" << endl;
std::this_thread::sleep_for(1s);
mtx.unlock();
cout << "\t deallocated " << counter << " device in thread" << endl;
pthread_exit(result);
}
};
void* threadTest1( void* result)
{
Derived dev;
dev.write(nullptr);
}
int main()
{
unsigned char byData[1024] = {0};
ssize_t len;
void *status = 0, *status2 = 0;
int result = 0, result2 = 0;
pthread_t pth, pth2;
pthread_create(&pth, NULL, threadTest1, &result);
pthread_create(&pth2, NULL, threadTest1, &result2);
//wait for all kids to complete
pthread_join(pth, &status);
pthread_join(pth2, &status2);
if (status != 0) {
printf("result : %d\n",result);
} else {
printf("thread failed\n");
}
if (status2 != 0) {
printf("result2 : %d\n",result2);
} else {
printf("thread2 failed\n");
}
return -1;
}
so the result is:
*Four or five arguments expected.
before Locking ... in thread
After Create 1 device in thread
before Locking ... in thread
After Create 2 device in thread
deallocated 2 device in thread
deallocated 2 device in thread
thread failed
thread2 failed
*
So here we can see that second thread comes to critical section before mutex was deallocated.
The string "After Create 2 device in thread" says about that.
If it comes to critical section before mutex is deallocated it means mutex works wrong.
If you have any thoughts please share.
thanks
The mutex itself is (probably) working fine (I'd recommend you to use std::lock_guard though), but both threads create their own Derived object, hence, they don't use the same mutex.
Edit: tkausl's answer is correct -- however, even if you switch to using a global mutex, the output may not change because of the detail in my answer so I'm leaving it here. In other words, there are two reasons why the output may not be what you expect, and you need to fix both.
Note in particular these two lines:
mtx.unlock();
cout << "\t deallocated " << counter << " device in thread" << endl;
You seem to be under the impression that these two lines will be run one right after the other, but there is no guarantee that this will happen in a preemptive multithreading environment. What can happen instead is that right after mtx.unlock() there could be a context switch to the other thread.
In other words, the second thread is waiting for the mutex to unlock, but the first thread isn't printing the "deallocated" message before the second thread preempts it.
The simplest way to get the output you expect would be to swap the order of these two lines.
You shall declare your mutex as a global variable and initiate it before calling pthread_create. You created two threads using pthread_create and both of them create their own mutex so there is absolutely no synchronization between them.
I have written this Producer/Consumer Problem solution. It seems to be working, other than the infinite loop. I was under the impression that pthread_exit(NULL); would make it stop, but honestly, I've become lost and confused. Could someone point me in the right direction of how to stop the loop?
#include<stdio.h>
#include<string.h>
#include<pthread.h>
#include<stdlib.h>
#include<unistd.h>
#include<iostream>
#include<semaphore.h>
#define BUFFSIZE 10
using namespace std;
int buffer[BUFFSIZE];
int size; //current buffer size
int n = 0, m = 0;
pthread_mutex_t Mutex = PTHREAD_MUTEX_INITIALIZER;
sem_t Available;
sem_t Buffer; //indicates if buffer is full
//----------------------------------------------------------------//
void *Consumers(void *argument)
{
int con_id = *((int *) argument);
while(1)
{
if(size == 0)
{
cout << "Queue is empty." << endl;
}
sem_wait(&Available);
pthread_mutex_lock(&Mutex);
size--;
cout << "Con " << con_id << ": Product removed from buffer" << endl;
//for(int i = 0; i < size; i++)
//{
// cout << Buffer[i] << " ";
//}
cout << endl;
pthread_mutex_unlock(&Mutex);
sem_post(&Buffer);
}
return(NULL);
}
//----------------------------------------------------------------//
void *Producers(void *argument)
{
int item = 8;
int pro_id = *((int *) argument);
while(1)
{
sem_wait(&Buffer);
pthread_mutex_lock(&Mutex);
//Buffer[size] = item;
cout << "Item added" << endl;
size++;
pthread_mutex_unlock(&Mutex);
sem_post(&Available);
}
return(NULL);
}
//----------------------------------------------------------------//
int main()
{
cout << "Enter number of producers: " << endl;
scanf("%d", &n);
cout << "Enter number of consumers: " << endl;
scanf("%d", &m);
//get number of producers(int n), and consumers(int m)
sem_init(&Available, 0, 0);
sem_init(&Buffer, 0, BUFFSIZE);
pthread_t *con = new pthread_t[m];
int *thread_args_c = new int[m];
for(int i = 0; i < n; i++)
{
thread_args_c[i] = i;
pthread_create(&con[i], NULL, Consumers, (void*) &i);
}
pthread_t *pro = new pthread_t[n];
int *thread_args_p = new int[n];
for(int i = 0; i < n; i++)
{
thread_args_p[i] = i;
pthread_create(&pro[i], NULL, Producers, (void*) &i);
pthread_join(con[i], NULL);
}
pthread_exit(NULL);
}
Not sure what you are expecting. pthread_exit appears in the end of the main (and completely not needed there, since main is exiting anyways), but your enless loops inside thread will never let main reach this point (since you are joining the consumers thread).
Also, your creation and joining model makes litle sense - what's the point of joining consumer thread after you've created a producer?
And last, but not the lease, you fail to join producer thread.
The loops will not stop because there is no logic in the code to actually exit the loop.
The process is stuck because pthread_join suspends the calling thread till the target exits. See documentation for pthread_join
If you don't care about actually terminating the threads and returning to the main thread, just remove the call to pthread_join. The process should terminate because the main thread exited.
To actually properly terminate the loops, you need to set an internal or external trigger. You could internally have the loops exit after a set number of iterations. For this, you will do while(x<=y) instead of while(1).
You could also make it more complicated and have the main thread signal the other threads externally that it wants the other threads to shut down. You can have the main thread set a (volatile) boolean when you are ready to exit and have the other threads break the loop based on it. If you care about the Atomicity of the exit, you will need to protect the boolean with a lock.
I need to implement barrier synchronization between 2 threads using mutex (only). Barrier synchronization is that 2 threads will wait for each other to meet at predefined step before proceeding.
I am able to do it using seamaphore but how can I achieve this only using mutex. I was given a hint that I need 2 mutex not 1 to do this.
Using Seamaphore:
#include <pthread.h>
#include <semaphore.h>
using namespace std;
sem_t s1;
sem_t s2;
void* fun1(void* i)
{
cout << "fun1 stage 1" << endl;
cout << "fun1 stage 2" << endl;
cout << "fun1 stage 3" << endl;
sem_post (&s1);
sem_wait (&s2);
cout << "fun1 stage 4" << endl;
}
void* fun2(void* i)
{
cout << "fun2 stage 1" << endl;
cout << "fun2 stage 2" << endl;
// sleep(5);
sem_post (&s2);
sem_wait (&s1);
cout << "fun2 stage 3" << endl;
}
main()
{
sem_init(&s1, 0, 0);
sem_init(&s2, 0, 0);
int value;
sem_getvalue(&s2, &value);
cout << "s2 = " << value << endl;
pthread_t iThreadId;
cout << pthread_create(&iThreadId, NULL, &fun2, NULL) << endl;
// cout << pthread_create(&iThreadId, NULL, &fun2, NULL) << endl;
pthread_create(&iThreadId, NULL, &fun1, NULL);
sleep(10);
}
Compile the above code as "g++ barrier.cc -lpthread"
How about NO MUTEXES and no locks? Using ATOMIC OPERATIONS only:
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <signal.h>
static sigset_t _fSigSet;
static volatile int _cMax=20, _cWait = 0;
static pthread_t _aThread[1000];
void * thread(void *idIn)
{
int nSig, iThread, cWait, id = (int)idIn;
printf("Start %d\n", id, cWait, _cMax);
// do some fake weork
nanosleep(&(struct timespec){0, 500000000}, NULL);
// barrier
cWait = __sync_add_and_fetch(&_cWait, 1);
printf("Middle %d, %d/%d Waiting\n", id, cWait, _cMax);
if (cWait < _cMax)
{
// if we are not the last thread, sleep on signal
sigwait(&_fSigSet, &nSig); // sleepytime
}
else
{
// if we are the last thread, don't sleep and wake everyone else up
for (iThread = 0; iThread < _cMax; ++iThread)
if (iThread != id)
pthread_kill(_aThread[iThread], SIGUSR1);
}
// watch em wake up
cWait = __sync_add_and_fetch(&_cWait, -1);
printf("End %d, %d/%d Active\n", id, cWait, _cMax);
return 0;
}
int main(int argc, char** argv)
{
pthread_attr_t attr;
int i, err;
sigemptyset(&_fSigSet);
sigaddset(&_fSigSet, SIGUSR1);
sigaddset(&_fSigSet, SIGSEGV);
printf("Start\n");
pthread_attr_init(&attr);
if ((err = pthread_attr_setstacksize(&attr, 16384)) != 0)
{
printf("pthread_attr_setstacksize failed: err: %d %s\n", err, strerror(err));
exit(0);
}
for (i = 0; i < _cMax; i++)
{
if ((err = pthread_create(&_aThread[i], &attr, thread, (void*)i)) != 0)
{
printf("pthread_create failed on thread %d, error code: %d %s\n", i, err, strerror(err));
exit(0);
}
}
for (i = 0; i < _cMax; ++i)
pthread_join(_aThread[i], NULL);
printf("\nDone.\n");
return 0;
}
I am not sure that you need two mutexes, with one mutex and a condition variable and an extra flag might be enough. The idea is that you enter the critical section by acquiring the mutex, then you check whether you are the first thread to come, if so, you wait on the condition. If you are the second thread coming then you wake up the waiting thread and both leave.