java - Netbeans，红色文本但没有错误

Question

我正在尝试为 Dinning Philosophers 实现一个解决方案。不确定我做得是否正确。我的程序没有崩溃，但输出中出现红色文本，但没有错误代码。

错误示例:

at (package_name).Phil.getrightFork(Phil.java:70) 第 70 行和第 46 行(调用 getrightFork 的行)之间的错误替代方案

我已经将 getrightFork 与 getleftFork 交换，但它总是选择 rightFork 作为错误

这是我正在使用的代码: 自定义信号量:

public class Semaphore {
public int value= 0 ;
public Semaphore(int value) {
    this.value = value;
}
public synchronized void up() { //notify must be syncrhonized
    value++;
    if (value > 0){
        this.notifyAll();
    }
}
public synchronized void down() throws InterruptedException {

    while (value <= 0){//Check if resource is avaiable, if not WAIT.
      this.wait();
    }
    value--; // Value is no longer negative
}}

主要:

public class main {

private static final int N = 10000;

public static void main(String[] args) throws InterruptedException{

    Phil[] phils = new Phil[N];
    Semaphore[] forks = new Semaphore[N];
    for (int i=0;i<N;i++){
      forks[i] = new Semaphore(1);
      phils[i] = new Phil(i, forks, N);
      phils[i].start();
    }
    for (int i=0;i<N;i++){
        phils[i].join();
    }
}}

菲尔类:

 public class Phil extends Thread {

  Semaphore fork[]; 
  int phil, total, left, right;
  boolean leftFork = false, rightFork = false;

public Phil(int spot ,Semaphore[] s, int N){

   phil = spot;
   left = spot;
   fork = s;
   switch(spot){
       case 0: 
           right = N-1;
               break;
       default:
           right = spot - 1;
           break;
   }

}

public void run(){

    System.out.println("I am Phil " + phil + " my left fork is " + left + " my right fork is " + right);
 while(true){
 try {
  if (phil%2 == 0){
      Thread.sleep(10); // Let the odd Phils eat first
  }   
  getrightFork();

  if (rightFork){
      getleftFork();
  }

  if (leftFork && rightFork){
      eat();
      retleftFork();
      retrightFork();
  }
  Thread.sleep(10);

 } catch (InterruptedException ex) {

 }
}
}
void getleftFork() throws InterruptedException{
    fork[left].down();
    //System.out.println("I got my left fork!");
    leftFork = true;
}
 void getrightFork() throws InterruptedException{
    fork[right].down();
    //System.out.println("I got my right fork!");
    rightFork = true;
}

void retleftFork(){
    fork[left].up();
    leftFork = false;

}

void retrightFork(){
    fork[right].up();
    rightFork = false;
}

void eat(){
    System.out.println("Phil:" + phil + " ate");
}}

Answer 1

您收到NullPointerException。您尝试访问的数组中的元素为 null。

这是由于您在整个数组完成之前启动 Phil 造成的...

Phil[] phils = new Phil[N];
Semaphore[] forks = new Semaphore[N];
for (int i = 0; i < N; i++) {
    forks[i] = new Semaphore(1);
    phils[i] = new Phil(i, forks, N);
    // Phil is starting, but how many phils are there??
    phil.start();
}

相反，尝试先填充数组，然后在单独的循环中启动它们...

Phil[] phils = new Phil[N];
Semaphore[] forks = new Semaphore[N];
for (int i = 0; i < N; i++) {
    forks[i] = new Semaphore(1);
    phils[i] = new Phil(i, forks, N);
}
for (Phil phil : phils)  {
    phil.start();
}
for (int i = 0; i < N; i++) {
    phils[i].join();
}

如果解决方案要求Phil在创建后立即启动，那么您需要更改检查代码以处理下一个元素可能为null的情况>