java中使用信号量的哲学家用餐

Question

我想用java信号量解决用餐哲学家的问题，但我被卡住了。最高ID的筷子应该是可用的，但它似乎总是采取，我不知道为什么。谁能告诉我我错在哪里了？

Fork类：

class Fork {
public static Semaphore fork = new Semaphore(1);
public int id;

Fork(int id) {
    this.id = id;
}

public int getId() {
    return id;
}

public boolean take() {
    return fork.tryAcquire();
}

public void putDown() {
    fork.release();
}}

哲学家班：

class Philosopher extends Thread {

private Fork fork_low;
private Fork fork_high;
private String name;

Philosopher(Fork fork_low, Fork fork_high, String name) {
    this.fork_low = fork_low;
    this.fork_high = fork_high;
    this.name = name;
}

public void run() {

    try {
        sleep(1000);
    } catch (InterruptedException ex) {
    }

    while (true) {
        eat();
    }
}

private void eat(){
    if(fork_low.take()){
        if(fork_high.take()){
            try {
                sleep(2000); // eating;
            } catch (InterruptedException ex) { }

            fork_high.putDown();
            fork_low.putDown();  

        }
        else{
            fork_low.putDown();
        }
    }
}}

主要内容：

public static void main(String[] args) {
    String[] names = {"Plato", "Aristotle", "Cicero", "Confucius", "Eratosthenes"};
    Fork[] fork = new Fork[5];
    Philosopher[] philosopher = new Philosopher[5];

    for (int i = 0; i < fork.length; i++) {
        fork[i] = new Fork(i);
    }

    for (int i = 0; i < philosopher.length; i++) {

        if (i != philosopher.length - 1) {
            philosopher[i] = new Philosopher(fork[i], fork[i+1], names[i]);
            philosopher[i].start();
        } else {
            philosopher[i] = new Philosopher(fork[0], fork[i], names[i]);
            philosopher[i].start();
        }
    }
}

Answer 1

这是用C语言解释解决的同一个问题

#include <stdio.h>
#include <pthread.h>
#include <semaphore.h>
#include <unistd.h>  
//if not used then gives warning for sleep used

//semaphore are basically designed to share the resources
// here the sem_t is the data type for the semaphore
sem_t room;//counting semaphore bcoz here only one instance of room butchair has 4
sem_t spoon[5]; //this is binary semaphore since every spoon has its own instance


void * philosopher(void *);
void eat(int);
int main()
{
int i;
int a[5];
pthread_t tid[5];// threads here are refrence to philosophers or diners bcoz we will have multiple users dining
                  
sem_init(&room,0,4);  
//1.pointer to declared semaphore
//2.pshared which has 0,1 value that is if 0 ->shared between threads                        
//                                      if 1 ->shared between process
//3.value with whch u initalise the semaphore
for(i=0;i<5;i++){
    //5 binary semaphore each for individual spoon
    sem_init(&spoon[i],0,1);
}

for(i=0;i<5;i++){
    a[i]=i;//allow 5 to enter at a time and deadlock occurs so let 4 of them in
    pthread_create(&tid[i],NULL,philosopher,(void*)&a[i]);
  //1.thread id 2.NULL 3.function 4.what you want to pass to the new thread   
  //here we pass the address of philosophers number to function
}

for(i=0;i<5;i++){
    pthread_join(tid[i],NULL);
}
}

void * philosopher(void * num){
int phil=*(int *)num; //cast the number passed as void to integer
//put sem_wait on both semaphore room and spoon
sem_wait(&room);//checks if resource is available,if then allocates and blocks semaphore
// room is counting semaphore so any is alocated then it decrements the count of total semaphore and 
// if all are allocated then it blocks thread and places it on queue untill resource is freed
printf("\nPhilospher number %d has sat on dining table\n",phil);
  
  sem_wait(&spoon[phil]);
  sem_wait(&spoon[(phil+1)%5]);
  //spoon is binary so if value of semaphore is 1 it is changed to 0 which means semaphore is allocated and cannot be used
  
  eat(phil);
  sleep(2);
  printf("\nPhilosopher %d has finished eating\n",phil);
  //free the semaphores so others in thread can use resources
  //returns +ve value on freeing semaphore
  //for binary semaphore if queue is empty then change semaphore value to 1 if not empty then remove process from queue and 
  // get it ready for allocation
  sem_post(&spoon[(phil+1)%5]);
  sem_post(&spoon[phil]);
  sem_post(&room);
}   

void eat(int phil){
  printf("\nPhilosopher %d is eating now\n",phil);
}

Answer 2

您有一个死锁，因为信号量在Fork类中是静态的，这相当于只有一个Fork可用。当您使信号量不是静态的（两个随机的哲学家同时运行）时，它可以完美地工作。

您可以观察在JDK的内置工具jvisualvm中工作的线程。