Question

它是一个简单的java线程。

有2个线程，应该每5秒打印一次没有错误。
但是我不能跑请帮我找出错误...

class MyThread extends Thread {
    String message;
    int delay;

    public MyThread(String s, int d) {
        message = s;
        delay = d;
    }
    public void run() {
        try {
            Thread.sleep(delay);
        } catch (InterruptedException e) {
            System.out.println(message + "+" + delay );
        } 
    }
}

public class applet {  
    public static void main(String[] args){  
        MyThread mt1, mt2;

        mt1 = new MyThread("Hello", 5000);
        mt2 = new MyThread("Bye  ", 10000);
        mt1.start();
        mt2.start();
   }  
}

Answer 1

如果发生异常，您正在打印 ONLY ONCE ...

将run方法修改为：

public void run() {
        try {
            Thread.sleep(delay);
            System.out.println(message + "+" + delay );
        } catch (InterruptedException e) {
            System.err.println("error here" );
        } 
    }

Answer 2

您的计划实际上有2个错误：

1。以正常流程打印您的信息

到目前为止，只有当你得到一个InterruptedException时才会打印你的消息，这是一个异常，当你的线程在休眠时被打断时会被抛出。因此，不应在catch块中打印您的消息，而是应该在sleep之后打印它：

try {
    Thread.sleep(delay);
    System.out.println(message + "+" + delay );
} catch (InterruptedException e) {
    // Re-set the interrupted flag
    Thread.currentThread().interrupt();
}

2。添加无限循环

由于您希望每5秒打印一次消息，您需要在无限循环中调用sleep，否则您的线程只打印一次并死掉，所以代码应该在最后：

try {
    while (true) {
        Thread.sleep(delay);
        System.out.printf("%s+%d%n", message, delay);
    }
} catch (InterruptedException e) {
    // Re-set the interrupted flag
    Thread.currentThread().interrupt();
}

3。奖金

或者，您可以使用ScheduledExecutorService使用方法scheduleAtFixedRate以固定费率安排任务，您的代码将是：

// Create a ScheduledExecutorService of 2 threads
ScheduledExecutorService executor = Executors.newScheduledThreadPool(2);
// Schedule my task every 5 seconds starting in 5 seconds
executor.scheduleAtFixedRate(new MyTask("Hello", 5000), 5L, 5L, TimeUnit.SECONDS);
// Schedule my task every 10 seconds starting in 10 seconds
executor.scheduleAtFixedRate(new MyTask("Bye  ", 10000), 10L, 10L, TimeUnit.SECONDS);

班级MyTask只是：

class MyTask implements Runnable {
    String message;
    int delay;

    public MyTask(String s, int d) {
        message = s;
        delay = d;
    }
    public void run() {
        System.out.printf("%s+%d%n", message, delay);
    }
}

java简单线程，但它无法运行

2 个答案:

1。以正常流程打印您的信息

2。添加无限循环

3。奖金