2019.9.10 6.828 barrier

╰+哭是因爲堅強的太久メ 2023-06-01 08:43 17阅读 0赞

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][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 <stdlib.h>
    #include <unistd.h>
    #include <stdio.h>
    #include <assert.h>
    #include <pthread.h>
    
    // #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

[https_pdos.csail.mit.edu_6.828_2018_homework_barrier.html]: https://pdos.csail.mit.edu/6.828/2018/homework/barrier.html