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！
------解决方案--------------------
Use the Event queue to retrieve event
- Java code
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(); }}
------解决方案--------------------
Queue data structure
- Java code
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);
}