我正在做一些多线程和Java并发功能的练习。我有 1 个生产者和 4 个消费者。现在我的问题是:当我确定生产者已经在 BlockingQueue 中完成生产时,还有其他更聪明的方法可以阻止消费者吗?现在我在队列中使用 -1 整数,但看起来非常基本!!谢谢
public class Exercise {
static class Producer implements Runnable {
int counter=0;
private BlockingQueue<Integer> queue;
Producer(BlockingQueue<Integer> q) {
queue = q;
}
public void run() {
try {
while (counter<100000000) {
queue.put(produce());
}
queue.put(new Integer(-1));
queue.put(new Integer(-1));
queue.put(new Integer(-1));
queue.put(new Integer(-1));
} catch (InterruptedException e) {
e.printStackTrace();
}
}
Integer produce() {
counter++;
return new Integer(counter);
}
}
static class Consumer implements Runnable {
private final BlockingQueue<Integer> queue;
private String name;
private long sum;
Consumer(BlockingQueue<Integer> q, String name) {
queue = q;
this.name=name;
sum=0;
}
public void run() {
try {
int x=0;
while (x>=0) {
x=queue.take();
if(x!=-1)sum+=x;
}
System.out.println(sum+" of "+ name);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
public static void main(String[] args) {
ExecutorService exec = Executors.newFixedThreadPool(6);
BlockingQueue<Integer> q =new LinkedTransferQueue<Integer>();
Producer p=new Producer(q);
Consumer c1 = new Consumer(q,"consumer1");
Consumer c2 = new Consumer(q,"consumer2");
Consumer c3 = new Consumer(q,"consumer3");
Consumer c4 = new Consumer(q,"consumer4");
exec.submit(p);
exec.submit(c1);
exec.execute(c2);
exec.submit(c3);
exec.execute(c4);
exec.shutdown();
}
}
您可以使用毒丸,但是使用这种毒丸的更有力方法是不要将其从队列中删除(或者如果您这样做,请将其放回原处) 这种生产者不需要知道你有多少消费者。
顺便说一句,我会避免使用明确的拳击,因为它更冗长、更慢。
而不是
queue.put(new Integer(-1));
你可以写
queue.put(-1);
甚至更好
static final int POISON_PILL = -1;
// on the producer
queue.put(POISON_PILL);
// on the consumer
while ((x = queue.take()) != POISON_PILL) {
sum += x;
queue.put(POISON_PILL);