2019.9.10 6.828 barrier

闲的蛋疼做了一个6.828的homework

就是这里 https://pdos.csail.mit.edu/6.828/2018/homework/barrier.html

就是说需要我们利用给的一个mutex，一个条件变量，实现一个barrier.

感觉barrier就是书上讲的那种sleep lock。具体到这个例子，前面说的mutex和条件变量，也就是下面的mutex和cond，分别就是一个spinlock和一个sleeplock。

pthread_cond_wait(&cond, &mutex);  // go to sleep on cond, releasing lock mutex
pthread_cond_broadcast(&cond);     // wake up every thread sleeping on cond

实际填充的内容在barrier{}里面，实际上就是控制bstate.nthread的值，每一个线程都检测一下这个值，来知道自己是第一个还是第二个进入线程的，如果是第一个的话就挂起来，是第二个的话就把第一个唤醒，这样来保证两个线程一起进入每个round。

我刚开始做的时候忘记每轮之后要round++了，困惑了好久。。。

#include 
#include 
#include 
#include 
#include 

// #define SOL

static int nthread = 1;
static int round = 0;

struct barrier {
  pthread_mutex_t barrier_mutex;
  pthread_cond_t barrier_cond;
  int nthread;      // Number of threads that have reached this round of the barrier
  int round;     // Barrier round
} bstate;




static void
barrier_init(void)
{
  assert(pthread_mutex_init(&bstate.barrier_mutex, NULL) == 0);
  assert(pthread_cond_init(&bstate.barrier_cond, NULL) == 0);
  bstate.nthread = 0;
}

static void 
barrier()
{ 
  pthread_mutex_lock(&bstate.barrier_mutex);
  if (!(bstate.nthread)) {
   bstate.nthread++; 
   pthread_cond_wait(&(bstate.barrier_cond), &(bstate.barrier_mutex)); 

}
else {
  bstate.nthread--;
  bstate.round++;
  pthread_cond_broadcast(&(bstate.barrier_cond));

}
 pthread_mutex_unlock(&bstate.barrier_mutex);
}

static void *
thread(void *xa)
{
  long n = (long) xa;
  long delay;
  int i;

  for (i = 0; i < 20000; i++) {
    int t = bstate.round;
    assert (i == t);
    barrier();
    usleep(random() % 100);
 
  }
}

int
main(int argc, char *argv[])
{
  pthread_t *tha;
  void *value;
  long i;
  double t1, t0;

  if (argc < 2) {
    fprintf(stderr, "%s: %s nthread\n", argv[0], argv[0]);
    exit(-1);
  }
  nthread = atoi(argv[1]);
  tha = malloc(sizeof(pthread_t) * nthread);
  srandom(0);

  barrier_init();

  for(i = 0; i < nthread; i++) {
    assert(pthread_create(&tha[i], NULL, thread, (void *) i) == 0);
  }
  for(i = 0; i < nthread; i++) {
    assert(pthread_join(tha[i], &value) == 0);
  }
  printf("OK; passed\n");
}

 

 

转载于:https://www.cnblogs.com/dynasty919/p/11495893.html