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我正在编写一个解决生产者/消费者问题的程序，特别是有界缓冲区版本（我相信它们的意思相同）。生产者将生成 x 个随机数，其中 x 是我的程序的命令行参数。目前，我相信我的程序正在进入一个无限循环，但我不确定它为什么会发生。我相信我正确地执行了信号量。

你像这样编译它： gcc -o prodcon prodcon.cpp -lpthread -lrt 然后运行，./prodcon 100（产生的随机数）

这是我的代码。

 typedef int buffer_item;

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

 #define BUFF_SIZE 10
 #define RAND_DIVISOR 100000000
 #define TRUE 1

 //two threads
 void *Producer(void *param);  
 void *Consumer(void *param);

 int insert_item(buffer_item item);
 int remove_item(buffer_item *item);
 int returnRandom();


 //the global semaphores

 sem_t empty, full, mutex;

 //the buffer

 buffer_item buf[BUFF_SIZE];

 //buffer counter
 int counter;

 //number of random numbers to produce
 int numRand;

 int main(int argc, char** argv) {
 /* thread ids and attributes */
 pthread_t pid, cid;  
 pthread_attr_t attr;
 pthread_attr_init(&attr);
 pthread_attr_setscope(&attr, PTHREAD_SCOPE_SYSTEM);

 numRand = atoi(argv[1]);
 sem_init(&empty,0,BUFF_SIZE);
 sem_init(&full,0,0);
 sem_init(&mutex,0,0);

 printf("main started\n");
 pthread_create(&pid, &attr, Producer, NULL);
 pthread_create(&cid, &attr, Consumer, NULL);
 printf("main gets here");
 pthread_join(pid, NULL);
 pthread_join(cid, NULL);
 printf("main done\n");

 return 0;

 }

 //generates a randum number between 1 and 100
 int returnRandom()
 {
  int num;
  srand(time(NULL));
  num = rand() % 100 + 1;
  return num;
 }

 //begin producing items
 void *Producer(void *param) {
    buffer_item item;
    int i;
    for(i = 0; i < numRand; i++)
    {
     //sleep for a random period of time
     int rNum = rand() / RAND_DIVISOR;
     sleep(rNum);

     //generate a random number
     item = returnRandom();

     //acquire the empty lock
     sem_wait(&empty);

     //acquire the mutex lock
     sem_wait(&mutex);

      if(insert_item(item)) 
      {
     fprintf(stderr, " Producer report error condition\n");
      }
      else 
      {
     printf("producer produced %d\n", item);
      }
      /* release the mutex lock */
     sem_post(&mutex);
     /* signal full */
     sem_post(&full);
    }
    return NULL;
 }

 /* Consumer Thread */
 void *Consumer(void *param) {
    buffer_item item;
    int i;
    for(i = 0; i < numRand; i++) {
    /* sleep for a random period of time */
    int rNum = rand() / RAND_DIVISOR;
    sleep(rNum);

    /* aquire the full lock */
    sem_wait(&full);
    /* aquire the mutex lock */
    sem_wait(&mutex);
    if(remove_item(&item)) {
    fprintf(stderr, "Consumer report error condition\n");
    }
    else {
    printf("consumer consumed %d\n", item);
    }
    /* release the mutex lock */
    sem_post(&mutex);
    /* signal empty */
    sem_post(&empty);
    }
    return NULL;
 }

  /* Add an item to the buffer */
 int insert_item(buffer_item item) {
    /* When the buffer is not full add the item
    and increment the counter*/
    if(counter < BUFF_SIZE) {
    buf[counter] = item;
    counter++;
    return 0;
    }
    else { /* Error the buffer is full */
    return -1;
    }
 }

 /* Remove an item from the buffer */
 int remove_item(buffer_item *item) {
    /* When the buffer is not empty remove the item
    and decrement the counter */
    if(counter > 0) {
    *item = buf[(counter-1)];
    counter--;
    return 0;
    }
    else { /* Error buffer empty */
    return -1;
    }
 }
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2 回答 2

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将互斥锁初始化为 1。尝试注释掉睡眠调用（我没有解释，除了“它对我有用”）。编辑：睡眠需要几秒钟，它可能看起来被阻塞，因为随机数太高。

在你的 main 中调用 srand （否则你会得到相同的值，因为你在同一时间再次播种）。

于 2010-04-12T18:20:03.763 回答
0 
 sem_init(&empty,0,BUFF_SIZE);
 sem_init(&full,0,0);

尝试在这里初始化“互斥”信号量。

于 2010-04-12T17:47:14.540 回答