如何把多个线程放到一个队列里？
需要产生很多线程。每个线程都执行一样的任务，比如调用某种资源。
不过不是每个线程都能顺利执行至结束的，所以没有成功的线程我想把它放到一个队列Queue里(比如用linkedlist来实现)，然后按照固定的时间，比如每5秒钟后，从队列里拿出最前面的那个线程再去执行当时失败的任务。
具体如何做才能实现这个过程那？
开始想到需要把失败的线程暂停，然后把线程对象放到队列里，等队列中前面的都执行完了，再重新激活这个线程再去尝试执行原来的任务。可是发现线程的suspend等都不能用。
那我可以在线程的RUN部分的代码里，如果发现执行失败，就在退出线程代码前加上 queue.addLast(this)么？这样能把这个线程对象加到队列里来么，等到再取出来时，会执行一样的功能？
或者有更好的解决方案？谢谢！
[解决办法]
Use the Event queue to retrieve event
import java.awt.AWTEvent;import java.awt.Container;import java.awt.EventQueue;import java.awt.Frame;import java.awt.Graphics;import java.awt.Point;import java.awt.Toolkit;import java.awt.event.ActionEvent;import java.awt.event.ActionListener;import java.awt.event.MouseEvent;import java.awt.event.WindowAdapter;import java.awt.event.WindowEvent;import javax.swing.JButton;import javax.swing.JFrame;import javax.swing.JPanel;public class EventQueuePanel extends JPanel implements ActionListener { EventQueuePanel() { JButton button = new JButton("Draw line"); add(button); button.addActionListener(this); } public void actionPerformed(ActionEvent evt) { Graphics g = getGraphics(); displayPrompt(g, "Click to chooose the first point"); Point p = getClick(); g.drawOval(p.x - 2, p.y - 2, 4, 4); displayPrompt(g, "Click to choose the second point"); Point q = getClick(); g.drawOval(q.x - 2, q.y - 2, 4, 4); g.drawLine(p.x, p.y, q.x, q.y); displayPrompt(g, "Done! Press button the start again."); g.dispose(); } public void displayPrompt(Graphics g, String s) { y += 20; g.drawString(s, 0, y); } public Point getClick() { EventQueue eq = Toolkit.getDefaultToolkit().getSystemEventQueue(); while (true) { try { AWTEvent evt = eq.getNextEvent(); if (evt.getID() == MouseEvent.MOUSE_PRESSED) { MouseEvent mevt = (MouseEvent) evt; Point p = mevt.getPoint(); Point top = getRootPane().getLocation(); p.x -= top.x; p.y -= top.y; return p; } } catch (InterruptedException e) { } } } private int y = 60; public static void main(String[] args) { JFrame frame = new JFrame(); frame.setTitle("EventQueueTest"); frame.setSize(300, 200); frame.addWindowListener(new WindowAdapter() { public void windowClosing(WindowEvent e) { System.exit(0); } }); Container contentPane = frame.getContentPane(); contentPane.add(new EventQueuePanel()); frame.show(); }}
[解决办法]
Queue data structure
public class Queue {
private int maxSize;
private long[] queArray;
private int front;
private int rear;
private int nItems;
public Queue(int s) {
maxSize = s;
queArray = new long[maxSize];
front = 0;
rear = -1;
nItems = 0;
}
// put item at end of a queue
public void insert(long j) {
if (rear == maxSize - 1) // deal with wraparound
rear = -1;
queArray[++rear] = j; // increment rear and insert
nItems++;
}
// take item from front of queue
public long remove() {
long temp = queArray[front++]; // get value and incr front
if (front == maxSize) // deal with wraparound
front = 0;
nItems--; // one less item
return temp;
}
public long peekFront() {
return queArray[front];
}
public boolean isEmpty() {
return (nItems == 0);
}
public boolean isFull() {
return (nItems == maxSize);
}
public int size() {
return nItems;
}
public static void main(String[] args) {
Queue theQueue = new Queue(5); // queue holds 5 items
theQueue.insert(10);
theQueue.insert(20);
theQueue.insert(30);
theQueue.insert(40);
theQueue.remove();
theQueue.remove();
theQueue.remove();
theQueue.insert(50);
theQueue.insert(60); // (wraps around)
theQueue.insert(70);
theQueue.insert(80);
while (!theQueue.isEmpty()) {
long n = theQueue.remove(); // (40, 50, 60, 70, 80)
System.out.print(n);
System.out.print(" ");
}
System.out.println("");
}
}
public class ThreadTest { public static void main(String args[]) { SimpleOne so = new SimpleOne("bzwm"); ThreadImpOne tio = new ThreadImpOne(so); new Thread(tio).start(); }}class SimpleOne { private transient int count = 0; private String name = null; public SimpleOne() { } public SimpleOne(String n) { setName(n); } public int getCount() { return count; } public void setCount(int count) { this.count = count; } public String getName() { return name; } public void setName(String name) { this.name = name; } public void test() throws Exception {//这里模拟任务失败，即第一次执行时失败。 if (count == 0) { count++; throw new Exception(); } else { System.out.println("hello " + name); } }}class ThreadImpOne implements Runnable { SimpleOne so = null; public ThreadImpOne(SimpleOne s) { so = s; } public void run() { try {//如果test方法抛出异常，则认为任务失败了。 so.test(); } catch (Exception e) { try { synchronized (so) { so.wait(3000); }//这里继续执行run(),直到可以成功，但在失败的时候，并没有让run方法返回 run(); } catch (InterruptedException e1) { } } }}
[解决办法]
LZ可不可以举一个具体的事例.....我想如果..一个线程失败了...你给他保存起来...下次再去运行.这个时候有可能这个线程还可能失败...如此反复...这样考虑的不周到把
[解决办法]
楼主应该重新思考一下设计，这跟队列没有什么关系。用一个 Vector 保存线程执行的参数，每启动一个线程就从中取出一个参数。线程执行失败的话，将参数重新放进 Vector 里面就可以了。
[解决办法]