我的 Produce Consumer 挂起
请复制下面的程序并尝试在您的 IDE 中运行。这是一个简单的 Produce Consumer 实现 - 当我使用一个 Producer 和一个 Consumer 线程时它运行良好，但当我使用 2 个线程时它会失败。请告诉我该程序挂起的原因或是否存在其他问题。
import java.util.LinkedList;
import java.util.Queue;
public class PCQueue {
private volatile Queue<Product> productQueue = new LinkedList<Product>();
public static void main(String[] args) {
PCQueue pc = new PCQueue();
Producer producer = new Producer(pc.productQueue);
Consumer consumer = new Consumer(pc.productQueue);
new Thread(producer, "Producer Thread 1").start();
new Thread(consumer, "Consumer Thread 1").start();
new Thread(producer, "Producer Thread 2").start();
new Thread(consumer, "Consumer Thread 2").start();
}
}
class Producer implements Runnable {
private Queue<Product> queue = null;
private static volatile int refSerialNumber = 0;
public Producer(Queue<Product> queue) {
this.queue = queue;
}
@Override
public void run() {
while (true) {
synchronized (queue) {
while (queue.peek() != null) {
try {
queue.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
queue.add(new Product(++refSerialNumber));
System.out.println("Produced by: "
+ Thread.currentThread().getName() + " Serial Number: "
+ refSerialNumber);
queue.notify();
}
}
}
}
class Consumer implements Runnable {
private Queue<Product> queue = null;
public Consumer(Queue<Product> queue) {
this.queue = queue;
}
@Override
public void run() {
while (true) {
synchronized (queue) {
while (queue.peek() == null) {
try {
queue.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
Product product = queue.remove();
System.out.println("Consumed by: "
+ Thread.currentThread().getName() + " Serial Number: "
+ product.getSerialNumber());
queue.notify();
}
}
}
}
class Product {
private int serialNumber;
public Product(int serialNumber) {
this.serialNumber = serialNumber;
}
public int getSerialNumber() {
return serialNumber;
}
}
Please copy the program below and try running in your IDE. It's a simple Produce Consumer implementation - it runs fine when I use one Producer and one Consumer thread but fails when using 2 each. Please let me know the reason why this program hangs or is there anything else wrong with it.
import java.util.LinkedList;
import java.util.Queue;
public class PCQueue {
private volatile Queue<Product> productQueue = new LinkedList<Product>();
public static void main(String[] args) {
PCQueue pc = new PCQueue();
Producer producer = new Producer(pc.productQueue);
Consumer consumer = new Consumer(pc.productQueue);
new Thread(producer, "Producer Thread 1").start();
new Thread(consumer, "Consumer Thread 1").start();
new Thread(producer, "Producer Thread 2").start();
new Thread(consumer, "Consumer Thread 2").start();
}
}
class Producer implements Runnable {
private Queue<Product> queue = null;
private static volatile int refSerialNumber = 0;
public Producer(Queue<Product> queue) {
this.queue = queue;
}
@Override
public void run() {
while (true) {
synchronized (queue) {
while (queue.peek() != null) {
try {
queue.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
queue.add(new Product(++refSerialNumber));
System.out.println("Produced by: "
+ Thread.currentThread().getName() + " Serial Number: "
+ refSerialNumber);
queue.notify();
}
}
}
}
class Consumer implements Runnable {
private Queue<Product> queue = null;
public Consumer(Queue<Product> queue) {
this.queue = queue;
}
@Override
public void run() {
while (true) {
synchronized (queue) {
while (queue.peek() == null) {
try {
queue.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
Product product = queue.remove();
System.out.println("Consumed by: "
+ Thread.currentThread().getName() + " Serial Number: "
+ product.getSerialNumber());
queue.notify();
}
}
}
}
class Product {
private int serialNumber;
public Product(int serialNumber) {
this.serialNumber = serialNumber;
}
public int getSerialNumber() {
return serialNumber;
}
}
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评论（2）
问题是您正在使用queue.notify()，它只会唤醒在队列上等待的单个线程。想象一下，生产者 1 调用 notification() 并唤醒生产者 2。生产者 2 看到队列中有东西，因此他不会生产任何东西，只是返回到 wait() 调用。现在，您的生产者和消费者都在等待通知，没有人需要通知任何人。
要解决代码中的问题，请使用queue.notifyAll() 唤醒在wait() 处阻塞的每个线程。这将允许您的消费者运行。
请注意，您的实现将队列限制为最多包含一项。因此，您不会从第二组生产者和消费者中看到任何好处。为了更好地全面实现，我建议您查看 BlockingQueue 并使用可以有界的实现，例如 ArrayBlockingQueue。无需同步和使用 wait/notify，只需使用 BlockingQueue.offer() 和 BlockingQueue.take()。
The problem is that you are using queue.notify() which will only wake up a single Thread waiting on the Queue. Imagine Producer 1 calls notify() and wakes up Producer 2. Producer 2 sees that there is something in the queue so he doesn't produce anything and simply goes back to the wait() call. Now both your Producers and Consumers are all waiting to be notified and nobody is left working to notify anyone.
To solve the problem in your code, use queue.notifyAll() to wake up every Thread blocked at a wait(). This will allow your consumers to run.
As a note, your implementation limits the queue to having at most one item in it. So you won't see any benefit from the second set of producers and consumers. For a better all around implementation, I suggest you look at BlockingQueue and use an implementation which can be bounded, for instance, the ArrayBlockingQueue. Instead of synchronizing and using wait/notify, simply use BlockingQueue.offer() and BlockingQueue.take().
使用queue.notifyAll()代替queue.notify()
instead of queue.notify() use queue.notifyAll()