我有一个要求任务异步执行,同时放弃任何进一步的请求,直到任务完成。单线程的任务,而不排队进一步的请求
同步该方法只是排队的任务,不跳过。我最初认为要使用SingleThreadExecutor,但它也排队完成任务。然后,我查看了ThreadPoolExecutor,但它读取队列以获取要执行的任务,因此将执行一个任务,并且排队最少一个任务(其他人可以使用ThreadPoolExecutor.DiscardPolicy放弃)。
我唯一能想到的就是使用Semaphore来阻塞队列。我用下面的例子来展示我想要达到的目标。有一种更简单的方法吗?我错过了明显的东西吗?
import java.util.concurrent.*;
public class ThreadPoolTester {
private static ExecutorService executor = Executors.newSingleThreadExecutor();
private static Semaphore processEntry = new Semaphore(1);
public static void main(String[] args) throws InterruptedException {
for (int i = 0; i < 20; i++) {
kickOffEntry(i);
Thread.sleep(200);
}
executor.shutdown();
}
private static void kickOffEntry(final int index) {
if (!processEntry.tryAcquire()) return;
executor.
submit(
new Callable<Void>() {
public Void call() throws InterruptedException {
try {
System.out.println("start " + index);
Thread.sleep(1000); // pretend to do work
System.out.println("stop " + index);
return null;
} finally {
processEntry.release();
}
}
}
);
}
}
样本输出
start 0
stop 0
start 5
stop 5
start 10
stop 10
start 15
stop 15
以axtavt答案,并转换上述例子给出了以下简单的解决方案。
import java.util.concurrent.*;
public class SyncQueueTester {
private static ExecutorService executor = new ThreadPoolExecutor(1, 1,
1000, TimeUnit.SECONDS,
new SynchronousQueue<Runnable>(),
new ThreadPoolExecutor.DiscardPolicy());
public static void main(String[] args) throws InterruptedException {
for (int i = 0; i < 20; i++) {
kickOffEntry(i);
Thread.sleep(200);
}
executor.shutdown();
}
private static void kickOffEntry(final int index) {
executor.
submit(
new Callable<Void>() {
public Void call() throws InterruptedException {
System.out.println("start " + index);
Thread.sleep(1000); // pretend to do work
System.out.println("stop " + index);
return null;
}
}
);
}
}
Coudl你告诉我,为什么你喜欢使用一个传播信号参考其他线程的堆栈DiscardPolicy捕获RejectedExecutionException? – 2012-10-16 01:06:11