java如何使多个线程并行运行,从而使线程一个接一个地执行
我正在尝试实现视差线程执行。比如说
t1, t2, and t3 are three threads , each performing a task to print multiplication table. I want to execute them in a sequence
--------------------------------
| t1 | t2 | t3 |
--------------------------------
| 2 | 3 | 4 |
| 4 | 6 | 8 |
| 6 | 9 | 12 |
| .. | .. | .. |
|------------------------------|
执行顺序:t1-->;t2-->;t3-->;t1
到目前为止,我能够创建三个独立执行任务的线程
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
public class SynchronizationExample1 {
public static void main(String[] args) {
ExecutorService es = Executors.newFixedThreadPool(3);
for(int i=1; i<=10; i++){
es.execute(new Task(i));
}
es.shutdown();
while(!es.isTerminated()){
}
System.out.println("finished all task");
}
}
任务类
public class Task implements Runnable {
private int i;
public Task(int i){
this.i = i;
}
@Override
public void run() {
System.out.println(Thread.currentThread().getName()+ "Start counter = " + i);
processMessage();
System.out.println(Thread.currentThread().getName()+ "End counter");
}
public void processMessage(){
try {
Thread.sleep(2000);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
输出:
pool-1-thread-1Start counter = 1
pool-1-thread-3Start counter = 3
pool-1-thread-2Start counter = 2
pool-1-thread-1End counter
pool-1-thread-1Start counter = 4
pool-1-thread-3End counter
pool-1-thread-3Start counter = 5
pool-1-thread-2End counter
pool-1-thread-2Start counter = 6
pool-1-thread-1End counter
pool-1-thread-1Start counter = 7
pool-1-thread-3End counter
pool-1-thread-3Start counter = 8
pool-1-thread-2End counter
pool-1-thread-2Start counter = 9
pool-1-thread-1End counter
pool-1-thread-1Start counter = 10
pool-1-thread-3End counter
pool-1-thread-2End counter
pool-1-thread-1End counter
finished all task
# 1 楼答案
这在术语上是矛盾的
你可以并行或按顺序做事,但不能同时做这两件事。(就像跑步和骑自行车同时进行!)
正如@Thilo所说,简单、实用(也是最好的)解决方案不是使用多个线程。只需使用一个线程和一个
for
循环如果您打算使用多个线程执行此操作,则需要线程进行同步。(这就是您当前代码的问题所在。您没有同步…只有三个“自由运行”线程和
sleep()
调用。这种方法永远不会可靠。)比如说,
直到你走到尽头。(那时你需要做一些特别的事情……)
对于每三对线程,该通知可以采用3个单独的“通道”(基本互斥体、锁存器、信号量、队列等)的形式,或者采用单个“通道”的形式,其中每个线程等待轮到它
(还有其他方法可以解决这个问题……但我们不需要在这里讨论。)