线程如何在java中工作

时间:2014-12-31 13:48:28

标签: java multithreading join

以下是我在某些方面感到困惑的代码。

public class SimpleThreads {

    // Display a message, preceded by
    // the name of the current thread
    static void threadMessage(String message) {
        String threadName =
            Thread.currentThread().getName();
        System.out.format("%s: %s%n",
                          threadName,
                          message);
    }

    private static class MessageLoop
        implements Runnable {
        public void run() {
            String importantInfo[] = {
                "Mares eat oats",
                "Does eat oats",
                "Little lambs eat ivy",
                "A kid will eat ivy too"
            };
            try {
                for (int i = 0;
                     i < importantInfo.length;
                     i++) {
                    // Pause for 4 seconds
                    Thread.sleep(4000);
                    // Print a message
                    threadMessage(importantInfo[i]);
                }
            } catch (InterruptedException e) {
                threadMessage("I wasn't done!");
            }
        }
    }

    public static void main(String args[])
        throws InterruptedException {

        // Delay, in milliseconds before
        // we interrupt MessageLoop
        // thread (default one hour).
        long patience = 1000 * 60 * 60;

        // If command line argument
        // present, gives patience
        // in seconds.
        if (args.length > 0) {
            try {
                patience = Long.parseLong(args[0]) * 1000;
            } catch (NumberFormatException e) {
                System.err.println("Argument must be an integer.");
                System.exit(1);
            }
        }

        threadMessage("Starting MessageLoop thread");
        long startTime = System.currentTimeMillis();
        Thread t = new Thread(new MessageLoop());
        t.start();

        threadMessage("Waiting for MessageLoop thread to finish");
        // loop until MessageLoop
        // thread exits
        while (t.isAlive()) {
            threadMessage("Still waiting...");
            // Wait maximum of 1 second
            // for MessageLoop thread
            // to finish.
            t.join(1000);
            if (((System.currentTimeMillis() - startTime) > patience)
                  && t.isAlive()) {
                threadMessage("Tired of waiting!");
                t.interrupt();
                // Shouldn't be long now
                // -- wait indefinitely
                t.join();
            }
        }
        threadMessage("Finally!");
    }
}

我需要解释的部分是     耐心等待= 1000 * 60 * 60;

耐心的目的是什么?在下面显示的代码部分,我认为根据条件获得另一个值。对?如果是这样,我们如何得到命令行参数(即args [0])?

if (args.length > 0) {
            try {
                patience = Long.parseLong(args[0]) * 1000;
            } catch (NumberFormatException e) {
                System.err.println("Argument must be an integer.");
                System.exit(1);
            }
        }
}

1 个答案:

答案 0 :(得分:2)

您在这里使用耐心变量:

if (((System.currentTimeMillis() - startTime) > patience)

现在System.currentTimeMillis() returns你当前的时间是毫秒。而你将耐心定义为

long patience = 1000 * 60 * 60;

假设你的args [0]你通过了1,你得到的最终耐心为3600000,相当于一小时,即60分钟* 60秒* 1000毫升

因此,如果你在一小时内阻塞并且线程处于活动状态,那么你等待线程无限期地完成,因为上面的条件满足使用join thread api。