用条件变量解决生产者和消费者问题（只有一个缓冲区）：

#define MAX 100                    //最大操作次数
 int buffer = 0;            //用来记录缓冲区中是否为空,只有一个缓冲区
Lock* mutex;
Condition* condc;
Condition* condp;
//生产者
void Producer(int tid)
{
 for(int i = 1;i < MAX;i++)
 {
    mutex->Acquire();            //加锁
    while(buffer != 0) condp->Wait(mutex);    //如果缓冲区值不为0，缓冲区满，等待消费者清空
    buffer = i;                //将缓冲区中值设为i
    printf("tid= %d,Produce a new item. i = %d\n",tid,i);
    condc->Signal(mutex);            //向消费者发送信号
    mutex->Release();            //解锁
 }
}
//消费者
void Consumer(int tid)
{
 for(int i = 1; i < MAX; i++)
 {
    mutex->Acquire();            //加锁
    while(buffer == 0) condc->Wait(mutex);    //当缓冲区值为0时，缓冲区为空，等待生产者
    buffer = 0;                //将缓冲区置为空
    printf("tid = %d,Consume a item. i = %d\n",tid,i);
    condp->Signal(mutex);            //向生产者发送信号
    mutex->Release();            //解锁
 }
}

void TestPC()
{
    mutex = new Lock("mutex");
    condc = new Condition("condc");
    condp = new Condition("condp");
    Thread *tp = Thread::getInstance("producer thread");
    if(tp != NULL) tp->Fork(Producer,tp->getTid());            //创建生产者线程
    Thread *tc = Thread::getInstance("consumer thread");
    if(tc != NULL) tc->Fork(Consumer,tc->getTid());            //创建消费者线程
}

结果如下：

利用信号量实现生产者于消费者问题，默认缓存区空间为10.List队列用来存放生产的产品。

#define MAXITEMS 10
List* list;
Semaphore* smutex;
Semaphore* full;
Semaphore* empty;
void Producer1(int tid)
{
    int* item;
    int flag = 1;
    for(int i = 0;i < 30;i++)
    {    
    item = &flag;                //生产一项
    printf("tid = %d, Produce a new item, item =%d\n",tid,(*item));
    empty->P();                //empty - 1
    smutex->P();
    list->Append(item);            //放入共享区域
    smutex->V();
    full->V();                //full + 1
    }  
}
void Consumer1(int tid)
{
        int* item;
         for(int i = 0;i < 30;i++)
        {
        full->P();                //full - 1
        smutex->P();
        item = (int*)list->Remove();        //从共享区域中移除
        smutex->V();
        empty->V();                //empty + 1
    *item = 0;                //消费一项
    printf("tid = %d, Consume a item, item =%d\n",tid,*item);
        }

}
void TestPC1()
{
    list = new List;
    smutex = new Semaphore("mutex",1);            //互斥信号量
    empty = new Semaphore("empty",MAXITEMS);        //记录空间中空余数
    full = new Semaphore("full",0);                //记录空间中存放数目
     Thread *tp = Thread::getInstance("producer thread");
        if(tp != NULL) tp->Fork(Producer1,tp->getTid());
        Thread *tc = Thread::getInstance("consumer thread");
        if(tc != NULL) tc->Fork(Consumer1,tc->getTid());
}

结果如下：