这是我的代码:
int BufferSize = 3;
int buffer[3] = {0,0,0};
int producer_cursor = 0;
int consumer_cursor = 0;
sem_t empty, mutex, full;
void* Producer(void *arg)
{
sem_wait(&empty);
sem_wait(&mutex);
printf("producer thread: #%d in critical setion\n", *((int*) arg));
int j;
for (j=0;j<BufferSize;j++) {
printf("%d", buffer[j]);
}
printf("\n");
buffer[producer_cursor] = 1;
producer_cursor = (++producer_cursor) % 3;
for (j=0;j<BufferSize;j++) {
printf("%d", buffer[j]);
}
printf("\n");
sem_post(&full);
sem_post(&mutex);
}
void * Consumer(void* arg)
{
sem_wait(&full);
sem_wait(&mutex);
printf("consumer thread: #%d in critical setion\n", *((int*) arg));
int j;
for (j=0;j<BufferSize;j++) {
printf("%d", buffer[j]);
}
printf("\n");
buffer[consumer_cursor] = 0;
consumer_cursor = (++consumer_cursor) % 3;
printf("after change\n");
for (j=0;j<BufferSize;j++) {
printf("%d", buffer[j]);
}
printf("\n");
sem_post(&empty);
sem_post(&mutex);
}
void main(int argc, char * argv[])
{
int n;
pthread_t *threads;
sem_init(&mutex, 0, 1);
sem_init(&empty, 0, 3);
sem_init(&full, 0, 0);
pthread_t consumers[NUM_COMSUMERS];
pthread_t producers[NUM_PRODUCERS];
int i;
for (i = 0; i < NUM_COMSUMERS; i++) {
pthread_create(consumers+i, NULL, Consumer, &i);
}
for (i = 0; i < NUM_PRODUCERS; i++) {
pthread_create(producers+i, NULL, Producer, &i);
}
for (i = 0; i < NUM_COMSUMERS; i++) {
pthread_join(consumers[i], NULL);
}
for (i = 0; i < NUM_PRODUCERS; i++) {
pthread_join(producers[i], NULL);
}
}
我得到的输出是:
producer thread: #1 in critical setion
000
100
consumer thread: #4 in critical setion
100
after change
000
producer thread: #0 in critical setion
000
010
consumer thread: #1 in critical setion
010
after change
000
producer thread: #1 in critical setion
000
001
producer thread: #2 in critical setion
001
101
consumer thread: #2 in critical setion
101
after change
100
consumer thread: #2 in critical setion
100
after change
000
producer thread: #2 in critical setion
000
010
consumer thread: #4 in critical setion
010
after change
000
结果是预期的,除了计数应该在0到4之间,而在这种情况下它没有“#3”。每次运行程序时,“#”都会改变。有时它甚至只显示“#0”和“#1”。
答案 0 :(得分:2)
pthread_create
的最后一个参数是问题。您将指针传递给i
,但i
的值会随着主循环的执行而更改。当您的线程执行打印并且您遵循指针时,无法保证i
仍然是0
(或1
或2
或3
)。< / p>
尝试这个简单的改变;它会给你预期的结果:
int i;
int id[4] = { 1, 2, 3, 4 };
for (i = 0; i < NUM_COMSUMERS; i++) {
pthread_create(consumers+i, NULL, Consumer, id + i);
}
for (i = 0; i < NUM_PRODUCERS; i++) {
pthread_create(producers+i, NULL, Producer, id + i);
}
以上假定NUM_CONSUMERS
和NUM_PRODUCERS
为4,需要重新设计以使其更通用。
或者,您没有必要传入指针,您可以执行以下操作:
for (i = 0; i < NUM_COMSUMERS; i++) {
pthread_create(consumers+i, NULL, Consumer, (void *)i);
}
for (i = 0; i < NUM_PRODUCERS; i++) {
pthread_create(producers+i, NULL, Producer, (void *)i);
}
在你的主题中,不要取消引用arg,只需将其转换为int
:
printf("producer thread: #%d in critical section\n", (int)arg);
和
printf("consumer thread: #%d in critical section\n", (int)arg);
注意:在64位系统上,指针和int之间会出现警告。