具有优先队列抛出异常的调度程序

Question

我试图运行具有优先级队列系统的调度程序的示例。它的工作方式如下：

使用包含优先级的GUI输入新进程。进程也通过GUI命令终止。上下文切换是通过命令来实现的，切换的原因并不重要。代码中使用了三个组件，其中包括：基于优先级的就绪队列，阻止列表和完成系统状态的输出，每个上下文切换显示就绪，已阻止和正在运行的进程。

import java.awt.Container; 
import java.awt.Dimension; 
import java.awt.FlowLayout; 
import java.awt.event.ActionEvent; 
import java.awt.event.ActionListener; 
import java.util.PriorityQueue; 
import java.util.Queue; 
import javax.swing.JButton; 
import javax.swing.JFrame; 
import javax.swing.JLabel; 
import javax.swing.JTextArea; 
import javax.swing.JTextField; 
class dispatcher extends JFrame implements ActionListener { 
    static int processNo = 0; 
    static Process currentlyRunning; 
    static Queue<Process> readyQueue; 
    static Queue<Process> blockedQueue; 
    static JButton newProcess; 
    static JButton contextSwitch; 
    static JTextField priority; 
    static JTextArea Queues; 

    private static Process newProcess(int i, int i0, String running) { 
     throw new UnsupportedOperationException("Not supported yet."); 
    } 
    dispatcher() { 
     Container cp = getContentPane(); 
     cp.setLayout(new FlowLayout()); 
     Queues = new JTextArea(5, 15); 
     Queues.setPreferredSize(new Dimension(500, 400)); 
     cp.add(Queues); 
     cp.add(new JLabel("Priority")); 
     priority = new JTextField(6); 
     newProcess = new JButton("Create Process"); 
     newProcess.addActionListener(this); 
     cp.add(priority); 
     cp.add(newProcess); 
     contextSwitch = new JButton("Context Switch"); 
     contextSwitch.addActionListener(this); 
     cp.add(contextSwitch); 
     setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); 
     setTitle("Dispatcher"); 
     setSize(600, 500); 
     setVisible(true); 
    } 
    public static void main(String args[]) { 
     dispatcher d = new dispatcher(); 
     readyQueue = new PriorityQueue<>(); 
     blockedQueue = new PriorityQueue<>(); 
     Process p1 = newProcess(processNo++, 2, "Running"); 
     Process p2 = newProcess(processNo++, 1, "Blocked"); 
     Process p3 = newProcess(processNo++, 4, "Ready"); 
     Process p4 = newProcess(processNo++, 3, "Ready"); 
     Process p5 = newProcess(processNo++, 5, "Ready"); 
     Process p6 = newProcess(processNo++, 8, "Blocked"); 
     readyQueue.add(p3); 
     readyQueue.add(p4); 
     readyQueue.add(p5); 
     blockedQueue.add(p6); 
     blockedQueue.add(p2); 
     currentlyRunning = p1; 
     contextSwitch(); 
    } 
    static void createProcess(int priority) { 
     readyQueue.add(newProcess(processNo++, priority, "ready")); 
    } 
    static void contextSwitch() { 

     String sb; 
     String currentState = null; 
     sb = " " + currentlyRunning 
       + "Currently Running Process"+"\n" 
       + "Process No  Priority Current State\n" 
       + currentlyRunning+processNo 
       + " "+currentlyRunning+priority+currentState + "\n\n"; 
     currentState = "Blocked"; 
     blockedQueue.add(currentlyRunning); 
     currentlyRunning = readyQueue.poll(); 
     currentState = "Running"; 
     sb = sb + "Ready Queue\n" + "Process No Priority Current State\n"; 
     for (Process p : readyQueue) { 
      sb = sb + p+processNo + " " + p+priority + " " + p+currentState 
        + "\n\n"; 
     } 
     sb = sb + "\nBlocked Queue\n" + "Process No Priority Current State\n"; 
     for (Process p : blockedQueue) { 
      sb = "\n" 
        + sb + p+processNo + " " + p+priority + " " + p+currentState; 
      Queues.setText(sb); 
     } 
    } 
    @Override 
    public void actionPerformed(ActionEvent e) { 
     if (e.getSource() == newProcess) { 
      System.out.print("process"); 
      int p = Integer.parseInt(priority.getText()); 
      p = (p < 0) ? 2 : p; 
      createProcess(p); 
     } else{ 
      System.out.print("context"); 
      contextSwitch(); 
     } 
    } 
} 

输出应该是这样的：click here 但我发现了一个空白的图形用户界面是这样的：click here 当我编译的代码，我得到了很多的例外。如果我输入一个数字并点击任何按钮，我就会收到未知来源。但是，如果没有在GUI中做任何事情，这是堆栈跟踪：

Exception in thread "main" java.lang.UnsupportedOperationException: Not supported yet. 
    at dispatcher.newProcess(dispatcher.java:24) 
    at dispatcher.main(dispatcher.java:50) 
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) 
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57) 
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) 
    at java.lang.reflect.Method.invoke(Method.java:606) 
    at com.intellij.rt.execution.application.AppMain.main(AppMain.java:144) 

Answer 1

你的代码是

public static void main(String args[]) { 
    dispatcher d = new dispatcher(); 
    readyQueue = new PriorityQueue<>(); 
    blockedQueue = new PriorityQueue<>(); 
    Process p1 = newProcess(processNo++, 2, "Running"); // <--- HERE 
    ... 

在线条为标志，你在呼唤你的方法newProcess()。该方法目前编码为抛出您看到的异常。

private static Process newProcess(int i, int i0, String running) { 
    throw new UnsupportedOperationException("Not supported yet."); 
} 

换句话说，您的代码完全符合您的要求。

要解决此问题，您必须编写代码为newProcess()。