请给出一段生产者消费者的代码使得满足以下三个条件
时间: 2024-02-17 21:02:26 浏览: 22
好的,以下是一段Java语言实现的生产者消费者代码,满足以下三个条件:
1. 使用信号量机制来实现同步和互斥操作;
2. 缓冲区大小为5,生产者生产10个产品,每个产品的生产时间为1秒,消费者消费所有产品;
3. 生产者和消费者交替执行,互不干扰。
```java
import java.util.concurrent.Semaphore;
public class ProducerConsumer {
private static final int BUFFER_SIZE = 5;
private static final int NUM_PRODUCTS = 10;
private static Semaphore mutex = new Semaphore(1);
private static Semaphore empty = new Semaphore(BUFFER_SIZE);
private static Semaphore full = new Semaphore(0);
private static int[] buffer = new int[BUFFER_SIZE];
private static int in = 0;
private static int out = 0;
public static void main(String[] args) {
Thread producer = new Thread(new Producer());
Thread consumer = new Thread(new Consumer());
producer.start();
consumer.start();
}
static class Producer implements Runnable {
public void run() {
for (int i = 0; i < NUM_PRODUCTS; i++) {
try {
empty.acquire();
mutex.acquire();
buffer[in] = 1;
System.out.println("Producer produces item " + (i + 1));
in = (in + 1) % BUFFER_SIZE;
} catch (InterruptedException e) {
e.printStackTrace();
} finally {
mutex.release();
full.release();
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
static class Consumer implements Runnable {
public void run() {
int sum = 0;
while (sum < NUM_PRODUCTS) {
try {
full.acquire();
mutex.acquire();
int item = buffer[out];
System.out.println("Consumer consumes item " + (sum + 1));
out = (out + 1) % BUFFER_SIZE;
sum++;
} catch (InterruptedException e) {
e.printStackTrace();
} finally {
mutex.release();
empty.release();
}
}
}
}
}
```
在这段代码中,我们使用了三个信号量:mutex、empty和full。其中,mutex用来实现互斥访问,empty用来表示缓冲区空闲的槽位数量,full用来表示缓冲区已经存储的产品数量。当empty的值为0时,生产者需要等待消费者消费产品;当full的值为0时,消费者需要等待生产者生产产品。通过这种方式,可以实现生产者和消费者之间的同步和互斥操作,避免资源竞争和死锁问题的发生。
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