CompletionService的poll方法

1、poll():马上返回完成的任务,若没有,则返回null

2、poll(long timeout, TimeUnit unit): 等待timeout时间,如果大于最短任务完成时间,则获取任务结果返回,结束等待;如果小于任务完成时间,则等待任务完成,获取结果并返回

实验代码:

import java.util.concurrent.Callable;
import java.util.concurrent.CompletionService;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.ExecutorCompletionService;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.TimeUnit;

public class MyPoll {

	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		ExecutorService executor=Executors.newCachedThreadPool();
		CompletionService<String> comservice=new ExecutorCompletionService<String>(executor);
		MyPollCallble callable=new MyPollCallble();
		MyPollCallble_time time_cable=new MyPollCallble_time();
		comservice.submit(callable);
		comservice.submit(time_cable);
		//poll():获取并移除表示下一个已完成任务的future。如果不存完成的任务,则返回null,即不会出现阻塞效果
//		System.out.println(comservice.poll());
		try {
			//等待指定timeout时间,在timeout之内获得值即向下执行,如果超时也向下执行
			System.out.println("get1  at"+System.currentTimeMillis());
			System.out.println("1 "+comservice.poll(1000, TimeUnit.SECONDS).get());
			System.out.println("get2  at"+System.currentTimeMillis());
			System.out.println("2 "+comservice.poll(2000, TimeUnit.SECONDS).get());
			System.out.println("get3  at"+System.currentTimeMillis());
		} catch (InterruptedException e) {
			// TODO 自动生成的 catch 块
			e.printStackTrace();
		} catch (ExecutionException e) {
			// TODO 自动生成的 catch 块
			e.printStackTrace();
		}
	}

}
class MyPollCallble implements Callable<String>{

	@Override
	public String call() throws Exception {
		// TODO 自动生成的方法存根
		System.out.println("start");
		System.out.println("first  "+System.currentTimeMillis());
		Thread.sleep(2000);
		System.out.println("end  "+System.currentTimeMillis());
		return "finish";
	}
	
}
class MyPollCallble_time implements Callable<String>{

	@Override
	public String call() throws Exception {
		// TODO 自动生成的方法存根
		System.out.println("start  time");
		System.out.println("second  "+System.currentTimeMillis());
		Thread.sleep(3000);
		System.out.println("end  time  "+System.currentTimeMillis());
		return "finish  time";
	}
	
}

  实验结果:

start
first  1492676279057
get1  at1492676279057
start  time
second  1492676279057
end  1492676281061
1 finish
get2  at1492676281061
end  time  1492676282064
2 finish  time
get3  at1492676282064

  可以看到get1与get2之间相差2秒,表明是在等待任务完成,而不是未获得值继续向下执行

 

posted @ 2017-04-20 16:04  L1019  阅读(1191)  评论(0编辑  收藏  举报