哲学家进餐问题3种代码

好的，以下是哲学家进餐问题的三种代码实现：

1. 使用死锁避免机制的实现：

import threading

class Chopstick:
    def __init__(self):
        self.lock = threading.Lock()

    def pick_up(self):
        self.lock.acquire()

    def put_down(self):
        self.lock.release()

class Philosopher(threading.Thread):
    def __init__(self, left_chopstick, right_chopstick):
        threading.Thread.__init__(self)
        self.left_chopstick = left_chopstick
        self.right_chopstick = right_chopstick

    def run(self):
        for i in range(1000):
            self.left_chopstick.pick_up()
            self.right_chopstick.pick_up()
            # 吃饭
            self.right_chopstick.put_down()
            self.left_chopstick.put_down()

if __name__ == "__main__":
    chopsticks = [Chopstick() for i in range(5)]
    philosophers = [Philosopher(chopsticks[i], chopsticks[(i+1)%5]) for i in range(5)]
    for philosopher in philosophers:
        philosopher.start()
    for philosopher in philosophers:
        philosopher.join()

2. 使用超时机制的实现：

import threading

class Chopstick:
    def __init__(self):
        self.lock = threading.Lock()

    def pick_up(self):
        self.lock.acquire()

    def put_down(self):
        self.lock.release()

class Philosopher(threading.Thread):
    def __init__(self, left_chopstick, right_chopstick):
        threading.Thread.__init__(self)
        self.left_chopstick = left_chopstick
        self.right_chopstick = right_chopstick

    def run(self):
        for i in range(1000):
            self.pick_up_chopsticks()
            # 吃饭
            self.put_down_chopsticks()

    def pick_up_chopsticks(self):
        while True:
            if self.left_chopstick.lock.acquire(timeout=1):
                if self.right_chopstick.lock.acquire(timeout=1):
                    return True
                else:
                    self.left_chopstick.lock.release()
            else:
                continue

    def put_down_chopsticks(self):
        self.left_chopstick.lock.release()
        self.right_chopstick.lock.release()

if __name__ == "__main__":
    chopsticks = [Chopstick() for i in range(5)]
    philosophers = [Philosopher(chopsticks[i], chopsticks[(i+1)%5]) for i in range(5)]
    for philosopher in philosophers:
        philosopher.start()
    for philosopher in philosophers:
        philosopher.join()

3. 使用信号量的实现：

import threading

class Chopstick:
    def __init__(self):
        self.lock = threading.Lock()

    def pick_up(self):
        self.lock.acquire()

    def put_down(self):
        self.lock.release()

class Philosopher(threading.Thread):
    def __init__(self, left_chopstick, right_chopstick, semaphore):
        threading.Thread.__init__(self)
        self.left_chopstick = left_chopstick
        self.right_chopstick = right_chopstick
        self.semaphore = semaphore

    def run(self):
        for i in range(1000):
            self.semaphore.acquire()
            self.left_chopstick.pick_up()
            self.right_chopstick.pick_up()
            # 吃饭
            self.right_chopstick.put_down()
            self.left_chopstick.put_down()
            self.semaphore.release()

if __name__ == "__main__":
    chopsticks = [Chopstick() for i in range(5)]
    semaphore = threading.Semaphore(4)
    philosophers = [Philosopher(chopsticks[i], chopsticks[(i+1)%5], semaphore) for i in range(5)]
    for philosopher in philosophers:
        philosopher.start()
    for philosopher in philosophers:
        philosopher.join()

以上三种实现方式都可以解决哲学家进餐问题，其中第三种使用信号量的实现是最常用和最简洁的实现方式。