Question

我正在尝试创建一个线程池，使工作人员和工作岗位得以创建，当他们忙碌时，不应该给他们的工作人员提供工作。到目前为止，我有这个代码，但是当我尝试运行它时，因为我有2个尝试并且在worker.java中运行run（）它不会按顺序输出它们。我希望第一次尝试发生，以便它显示所有工人已经开始工作，然后我想让它说出哪个工人开始工作。

如果有办法使用busy = true;说这个工人很忙，所以把工作交给下一个工人。

Worker.java

public class Worker implements Runnable {

    private final int id;  // Unique worker ID.

    private final JobStack jobStack;  // Reference to the job stack.
    private final ResourceStack resourceStack;  // Reference to the resource stack.

    private Job job;  // Job being processed.
    private Resource[] resources;  // Resources being used for job being processed.

    private boolean busy;  // Indicates the status of the worker. True when they are working (executing jobs) and false when there are no more jobs left to execute.

    private final Map<Integer, ArrayList<Integer>> jobsCompleted;  // The job record of the worker. Stores each job's ID and the IDs of the resources used for each job.

    // Constructor.
    public Worker(int theId, JobStack theJobStack, ResourceStack theResourceStack) {
        id = theId;
        jobStack = theJobStack;
        resourceStack = theResourceStack;
        job = null;
        busy = true;
        jobsCompleted = new TreeMap<>();
    }

    public void run() {
        try
        {
            System.out.println("Worker " + id + " started Work ");
            Thread.sleep(100);
        }    
        catch (Exception e)
        {            
            System.out.println ("Exception is caught");
        }

        for (Job job = jobStack.pop(); job != null; job = jobStack.pop()) {
            try{
                System.out.println("Worker " + id + " started Job "  + job.getId());
            }
            catch (Exception e)
            {            
                System.out.println ("Exception is caught");
            }
        }
     }
 }

Workforce.java

import java.util.logging.Level;
import java.util.logging.Logger;

public class Workforce {

    private final Worker[] pool;  // The worker population.
    private int workerCount = 0;  // Used to generate each worker's ID and to keep a record of the number of workers in the workforce.

    Thread[] workerThreads;

    private final JobStack jobStack;  // Reference to the job stack.
    private final ResourceStack resourceStack;  // Reference to the resource stack.

    // Constructor.
    public Workforce(int size, JobStack theJobStack, ResourceStack theResourceStack) {
        jobStack = theJobStack;
        resourceStack = theResourceStack;

        pool = new Worker[size];
        for(int i=0; i<pool.length; i++) {
            pool[i] = new Worker(workerCount, jobStack, resourceStack);
            workerCount++;
        }

        workerThreads = new Thread[pool.length];
        for(int i=0; i<workerThreads.length; i++) {
            workerThreads[i] = new Thread(pool[i]);
        }
    }


    /// UNDER CONSTRUCTION /////////////////////////////////////////////////////


    // Starts all the worker threads.
    public void start() {

        for(int i=0; i<pool.length; i++) {
           workerThreads[i] = new Thread(pool[i]);           
            workerThreads[i].start();}
    }



    // Checks whether all workers have finished.
    public boolean allWorkersFinished() {

       return false;
    }
    // Prints the job record of all workers.
    public void printJobRecords() {
        ;
    }
}

Answer 1

您需要在第一个和第二个try-catch块之间暂停工作线程，并在管理器完成初始化后再次恢复。您可以通过wait()/notify()实现此目的，这些wait()旨在阻止（释放）一个线程，直到满足特定条件为止。

我的设计在我看来基于 manager-worker（主从）并行架构。经理在开始时启动所有工人，之后经理告诉所有工人开始工作。在工作方面，在初始化之后，它立即切换到暂停状态并等待来自管理器的下一个命令。

在经理 - 工作者模型中，通常，经理维护工作池，一组工作项，并且根据请求，经理将下一个可用工作项发送给工人。每个工人一次处理一个工作项，完成后，请求经理提供新的工作项目。这一直持续到没有剩余工作项目为止。当所有工作项目完成后，经理会告诉工人停止工作。

一种可能的解决方案是在经理和工人之间共享一个共同的锁。在您的第一个try-catch块之后，所有工作人员在该锁上调用notifyAll()，并且当所有工作程序初始化时，管理器在同一个锁上调用final Object lock = new Object();。

boolean isWorking = false;

// your first try-catch block

synchronized (lock) {
    try {
        while (!isWorking) {
            lock.wait();
        }

        isWorking = true;
    } 
    catch (Exception e) {
            e.printStackTrace();            
    } 
}

// your second try-catch block

// do your work here!

<强> Wroker

// Starts all the worker threads.  

synchronized (lock) {
    lock.notifyAll(); // notify workers that initialization is done!
}

<强>管理器

find . -name "*.mp4" -print0 | xargs -0 -I $ bash -c "ffmpeg -v error -xerror -i $ -f null - || echo $"

Threadpool并尝试捕获到下一个

1 个答案: