在 C 中使用互斥锁和屏障进行线程同步

Question

我正在尝试使用 pthread 互斥变量和屏障来同步我的程序的输出，但它没有按照我想要的方式工作。每个线程每 20 个值（来自 for 循环）看到它的最终值，这没关系，但我试图让它们都达到相同的最终值（如果使用 5 个线程，它们都应该看到 100 作为最终值，有 4 个线程，80 个等）

#include <pthread.h>
#include <stdio.h>
#include <stdlib.h> 

pthread_mutex_t mutex1 = PTHREAD_MUTEX_INITIALIZER;

int SharedVariable =0;
void *SimpleThread(void *args)
{
    int num,val,rc;
    int which =(int)args;
    rc = pthread_mutex_lock(&mutex1);
    for(num=0; num<20; num++){
#ifdef PTHREAD_SYNC
        if(random() > RAND_MAX / 2)
            usleep(10);
#endif
        //pthread_mutex_lock(&mutex1);
        val = SharedVariable;
        printf("*** thread %d sees value %d\n", which, val);
        //pthread_mutex_lock(&mutex1);
        SharedVariable = val+1;
        pthread_mutex_unlock(&mutex1);
    }   
    val=SharedVariable;

    printf("Thread %d sees final value %d\n", which, val);
    //pthread_mutex_destroy(&mutex1);
    //pthread_exit((void*) 0);
    //pthread_mutex_unlock(&mutex1);

}

int main (int argc, char *argv[])
{
   if(atoi(argv[1]) > 0){        
   int num_threads = atoi(argv[1]);  
   //pthread_mutex_init(&mutex1, NULL);
   pthread_t threads[num_threads];
   int rc;
   long t;
   rc = pthread_mutex_lock(&mutex1);
   for(t=0; t< num_threads; t++){
      printf("In main: creating thread %ld\n", t);
      rc = pthread_create(&threads[t], NULL, SimpleThread, (void* )t);
      if (rc){
         printf("ERROR; return code from pthread_create() is %d\n", rc);
         exit(-1);
      }

    //pthread_join(thread1);

   }
    rc= pthread_mutex_unlock(&mutex1);
}
   else{
      printf("ERROR: The parameter should be a valid positive number.");
      exit(-1);
  }

   pthread_mutex_destroy(&mutex1);
   pthread_exit(NULL);
}

非常感谢任何建议或帮助！ 提前致谢！

Answer 1

您需要在检查最终值之前使用屏障 (pthread_barrier_wait()) - 这可确保在所有线程都到达屏障之前不会继续执行任何线程。

此外，您应该调用pthread_join() 来等待线程完成，并且您只需要在增量周围保持互斥锁：

#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

pthread_mutex_t mutex1 = PTHREAD_MUTEX_INITIALIZER;
pthread_barrier_t barrier1;

int SharedVariable = 0;

void *SimpleThread(void *args)
{
    int num,val;
    int which = (int)args;

    for(num = 0; num < 20; num++) {
#ifdef PTHREAD_SYNC
        if(random() > RAND_MAX / 2)
            usleep(10);
#endif
        pthread_mutex_lock(&mutex1);
        val = SharedVariable;
        printf("*** thread %d sees value %d\n", which, val);
        SharedVariable = val + 1;
        pthread_mutex_unlock(&mutex1);
    }

    pthread_barrier_wait(&barrier1);

    val = SharedVariable;
    printf("Thread %d sees final value %d\n", which, val);
    return 0;
}

int main (int argc, char *argv[])
{
    int num_threads = argc > 1 ? atoi(argv[1]) : 0;

    if (num_threads > 0) {
        pthread_t threads[num_threads];
        int rc;
        long t;

        rc = pthread_barrier_init(&barrier1, NULL, num_threads);

        if (rc) {
            fprintf(stderr, "pthread_barrier_init: %s\n", strerror(rc));
            exit(1);
        }

        for (t = 0; t < num_threads; t++) {
            printf("In main: creating thread %ld\n", t);
            rc = pthread_create(&threads[t], NULL, SimpleThread, (void* )t);
            if (rc) {
                printf("ERROR; return code from pthread_create() is %d\n", rc);
                exit(-1);
            }
        }

        for (t = 0; t < num_threads; t++) {
            pthread_join(threads[t], NULL);
        }
    }
    else {
        printf("ERROR: The parameter should be a valid positive number.\n");
        exit(-1);
    }

    return 0;
}

Answer 2

尝试将pthread_mutext_unlock(&mutext1) 移出SimpleThread 中的for 循环。您在原始代码中锁定一次并解锁多次（20）次。

或者，您可以将 pthread_mutex_lock(&mutext1) 移动到 for 循环中，就在您阅读和修改您的 SharedVariable 之前。在这种情况下，每个线程的add-by-one操作可能不连续，但每个线程都会得到正确的最终值。

在读取SharedVariable 的最终值之前，请使用屏障等待所有线程完成它们的工作。