我已經編寫了一個解決方案,標準的生產者 - 消費者問題使用一個大小爲5的緩衝區和pthreads與一個空的和完整的信號量和互斥鎖。我認爲一切都按預期工作,但只是注意到我得到堆棧行爲(LIFO)而不是預期的隊列(FIFO)行爲。我已經搜索,但無法找到任何類似的問題,因爲除了訂單以外,我正按預期生產和消費。生產者 - 消費者堆棧行爲,而不是隊列
這是一項家庭作業,所以我沒有真正尋找代碼我只是想知道在哪裏尋找錯誤或知道爲什麼行爲可能會比預期的不同。
struct data
{
pthread_mutex_t mutex;
sem_t full;
sem_t empty;
};
int bufferCount;
buffer_item buffer[BUFFER_SIZE];
pthread_t thread;
int insert_item(buffer_item item)
{
if (bufferCount < BUFFER_SIZE)
{
buffer[bufferCount] = item;
++bufferCount;
return 0;
}
else
return -1; //insert failed
}
int remove_item(buffer_item *item)
{
if (bufferCount > 0)
{
*item = buffer[bufferCount - 1];
--bufferCount;
return 0;
}
else
return -1; //error failed to remove
}
void Initialize (void *param)
{
struct data *locks = param;
pthread_mutex_init(&(locks->mutex), NULL);
sem_init(&(locks->full), 0, 0);
sem_init(&(locks->empty), 0, BUFFER_SIZE);
bufferCount = 0;
}
void *producer (void *param)
{
struct data *locks = param;
do
{
sleep(rand()%5 + 1); //sleep for between 1 and 5 seconds
buffer_item num = rand();
sem_wait(&(locks->empty));
pthread_mutex_lock(&(locks->mutex));
if (insert_item(num))
{
printf("Insert in producer failed\n");
exit(1);
}
else
printf("Producer produced %d\n", num);
pthread_mutex_unlock(&(locks->mutex));
sem_post(&(locks->full));
}while(1);
}
void *consumer (void *param)
{
struct data *locks = param;
do
{
sleep(rand()%5 + 1); //sleep for between 1 and 5 seconds
buffer_item num;
sem_wait(&(locks->full));
pthread_mutex_lock(&(locks->mutex));
if (remove_item(&num))
{
printf("Remove in consumer failed\n");
exit(1);
}
else
printf("Consumer consumed %d\n", num);
pthread_mutex_unlock(&(locks->mutex));
sem_post(&(locks->empty));
}while(1);
}
int main(int argc, char *argv[])
{
if (argc != 4)
{
printf("Incorrect number of arguments should be 4\n");
exit (1);
}
int sleepTime = atoi(argv[1]);
int producerThreads = atoi(argv[2]);
int consumerThreads = atoi(argv[3]);
struct data *locks = (struct data *) malloc(sizeof(struct data));
Initialize(locks);
for (int i =0; i < producerThreads; ++i)
pthread_create(&thread, NULL, producer, locks);
for(int i = 0; i < consumerThreads; ++i)
pthread_create(&thread, NULL, consumer, locks);
sleep(sleepTime);
free (locks);
return 0;
}
錯誤是在第42行 –
有趣.....我確信我可以發佈代碼和數百人可以使它更好,並修復它,但我想從中吸取教訓。因此,一般來說,在生產者消費者問題中,可能會影響消費者從緩衝區中移除的訂單。 – user2514231
你對我們的期望是什麼?如果您編寫的是LIFO而不是FIFO,那麼顯然您的消費者會從容器錯誤的端點消耗掉。但這很明顯,你一定要自己檢查這個 –