Question

如果方法本身需要一定的时间（少于x秒）运行，我是否知道安排每隔x秒运行一次方法的最佳方法是什么？

例如：

import java.util.Timer;
import java.util.TimerTask;
import java.util.concurrent.TimeUnit;

class Helper extends TimerTask 
{ 
    public void run() 
    { 
        System.out.println("Start");
        //for example, the program takes about 100 ms to run and the exact time may vary
        try {
            TimeUnit.MILLISECONDS.sleep(100);
        } catch (InterruptedException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        System.out.println("End");

    } 
} 

public class Test 
{ 
    public static void main(String[] args) 
    { 

        Timer timer = new Timer(); 
        TimerTask task = new Helper();
        timer.schedule(task, 0, 1000); 

    } 
}

如何保证在0s之后第二次调度函数是1s（没有延迟）而不是1.1s？

Answer 1

要在上一个任务结束后的毫秒数开始任务，可以使用ScheduledExecutorService：

import java.util.concurrent.Executors;
import java.util.concurrent.ScheduledExecutorService;
import java.util.concurrent.TimeUnit;

public class Test
{
    public static void main(String[] args)
    {
        ScheduledExecutorService seService = Executors.newSingleThreadScheduledExecutor();
        Runnable task = new Helper(seService);
        seService.execute(task);
    }
}

class Helper implements Runnable
{
    private final ScheduledExecutorService seService;

    Helper(ScheduledExecutorService seService) {
        this.seService = seService;
    }
    @Override
    public void run()
    {
        System.out.println("Start");

        try {
            TimeUnit.MILLISECONDS.sleep(100);
        } catch (InterruptedException e) {     e.printStackTrace();  }

        seService.schedule(this, 1000, TimeUnit.MILLISECONDS); //execute again after delay
        System.out.println("End");
    }
}

Answer 2

像这样更改代码：

import java.util.Timer;
import java.util.TimerTask;
import java.util.concurrent.TimeUnit;

class Helper extends TimerTask 
{ 
    public void run() 
    { 
        long current = System.currentTimeMillis();
        System.out.println("Start");
        //for example, the program takes about 100 ms to run and the exact time may vary
        try {
            TimeUnit.MILLISECONDS.sleep(100);
        } catch (InterruptedException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        System.out.println("End");
        long after = System.currentTimeMillis();
        Timer timer = new Timer(); 
        TimerTask task = new Helper();
        timer.schedule(task, 0, 1000 - (after - current)); 

    } 
} 

public class Test 
{ 
    public static void main(String[] args) 
    { 

        Timer timer = new Timer(); 
        TimerTask task = new Helper();
        timer.schedule(task, 0, 1000); 
    } 
}

另一个选择是使用thread.sleep（）：

import java.util.Timer;
import java.util.TimerTask;
import java.util.concurrent.TimeUnit;

class Helper extends TimerTask 
{ 
    public void run() 
    { 
        while(true)
        {
            long current = System.currentTimeMillis();
            System.out.println("Start");
            //for example, the program takes about 100 ms to run and the exact time may vary
            try {
                TimeUnit.MILLISECONDS.sleep(100);
            } catch (InterruptedException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            System.out.println("End");
            long after = System.currentTimeMillis();
            Thread.sleep(1000 - (after - current)); 
        }

    } 
} 

public class Test 
{ 
    public static void main(String[] args) 
    { 

        Timer timer = new Timer(); 
        TimerTask task = new Helper();
        timer.schedule(task, 0, 1000); 
    } 
}

如果方法本身需要一些时间，我们如何仍每X秒运行一个方法？

2 个答案: