Executors.newFixedThreadPool和ArrayBlockingQueue一点使用心得

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kaixin2010 2016-05-11 15:05:24

文章标签 Executors.newFixedTh 文章分类 Java 后端开发

http://heipark.iteye.com/blog/1393847

newFixedThreadPool内部有个任务队列，假设线程池里有3个线程，提交了5个任务，那么后两个任务就放在任务队列了，即使前3个任务sleep或者堵塞了，也不会执行后两个任务，除非前三个任务有执行完的newFixedThreadPool使用范例：

Java代码

import java.io.IOException;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
public class Test {
public static void main(String[] args) throws IOException, InterruptedException {
ExecutorService service = Executors.newFixedThreadPool(2);
for (int i = 0; i < 6; i++) {
final int index = i;
System.out.println("task: " + (i+1));
Runnable run = new Runnable() {
@Override
public void run() {
System.out.println("thread start" + index);
try {
Thread.sleep(Long.MAX_VALUE);
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println("thread end" + index);
}
};
service.execute(run);
}
}
}

输出：

task: 1
task: 2
thread start0
task: 3
task: 4
task: 5
task: 6
task: 7
thread start1
task: 8
task: 9
task: 10
task: 11
task: 12
task: 13
task: 14
task: 15

从实例可以看到for循环并没有被固定的线程池阻塞住，也就是说所有的线程task都被提交到了ExecutorService中，查看 Executors.newFixedThreadPool()如下：

public static ExecutorService newFixedThreadPool(int nThreads) {
return new ThreadPoolExecutor(nThreads, nThreads,
0L, TimeUnit.MILLISECONDS,
new LinkedBlockingQueue<Runnable>());
}

可以看到task被提交都了LinkedBlockingQueue中。这里有个问题，如果任务列表很大，一定会把内存撑爆，如何解决？看下面：

Java代码

import java.io.IOException;
import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.BlockingQueue;
import java.util.concurrent.ThreadPoolExecutor;
import java.util.concurrent.TimeUnit;
public class Test {
public static void main(String[] args) throws IOException, InterruptedException {
BlockingQueue<Runnable> queue = new ArrayBlockingQueue<Runnable>(3);
ThreadPoolExecutor executor = new ThreadPoolExecutor(3, 3, 1, TimeUnit.HOURS, queue, new ThreadPoolExecutor.CallerRunsPolicy());
for (int i = 0; i < 10; i++) {
final int index = i;
System.out.println("task: " + (index+1));
Runnable run = new Runnable() {
@Override
public void run() {
System.out.println("thread start" + (index+1));
try {
Thread.sleep(Long.MAX_VALUE);
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println("thread end" + (index+1));
}
};
executor.execute(run);
}
}
}