我在一本书上发现了 Java 中“餐饮哲学家问题”的替代解决方案：
public class Philosopher extends Thread {
private final int maxPause = 100;
private int bites = 10;
private Chopstick lower;
private Chopstick higher;
private int index;
public Philosopher(int i, Chopstick left, Chopstick right) {
index = i;
if (left.getNumber() < right.getNumber()) {
this.lower = left;
this.higher = right;
} else {
this.lower = right;
this.higher = left;
}
}
public void eat() {
System.out.println("Philosopher " + index + ": start eating");
pickUp();
chew();
putDown();
System.out.println("Philosopher " + index + ": done eating");
}
public void pickUp() {
pause();
lower.pickUp();
pause();
higher.pickUp();
pause();
}
public void chew() {
System.out.println("Philosopher " + index + ": eating");
pause();
}
public void pause() {
try {
int pause = AssortedMethods.randomIntInRange(0, maxPause);
Thread.sleep(pause);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
public void putDown() {
higher.putDown();
lower.putDown();
}
public void run() {
for (int i = 0; i < bites; i++) {
eat();
}
}
}
public class Chopstick {
private Lock lock;
private int number;
public Chopstick(int n) {
lock = new ReentrantLock();
this.number = n;
}
public void pickUp() {
lock.lock();
}
public void putDown() {
lock.unlock();
}
public int getNumber() {
return number;
}
}
解决方案的文本是：
或者，我们可以用从 e 到 N - 1 的数字标记筷子。每个哲学家都试图先拿起编号较低的筷子。这实质上意味着每个哲学家都先选择左筷子，然后再选择右筷子（假设这是您标记它的方式），除了最后一位相反的哲学家。有了这个解决方案，哲学家就永远不能不拿着较小的筷子就拿着较大的筷子。这阻止了具有循环的能力，因为循环意味着较高的筷子将“指向”较低的筷子。
但对我来说还不清楚。有人可以帮我举个例子吗？
谢谢
- - 编辑 - - -
主类：
public class Question {
public static int size = 3;
public static int leftOf(int i) {
return i;
}
public static int rightOf(int i) {
return (i + 1) % size;
}
public static void main(String[] args) {
Chopstick[] chopsticks = new Chopstick[size + 1];
for (int i = 0; i < size + 1; i++) {
chopsticks[i] = new Chopstick(i);
}
Philosopher[] philosophers = new Philosopher[size];
for (int i = 0; i < size; i++) {
Chopstick left = chopsticks[leftOf(i)];
Chopstick right = chopsticks[rightOf(i)];
philosophers[i] = new Philosopher(i, left, right);
}
for (int i = 0; i < size; i++) {
philosophers[i].start();
}
}
}