Python で 4 種類のロックを使用する例 (コード)

num = 0 def a(): global num for _ in range(10000000): num += 1 def b(): global num for _ in range(10000000): num += 1 if __name__ == '__main__': t1=Thread(target=a) t1.start() t2=Thread(target=b) t2.start() t1.join() t2.join() print(num) #基本永远会小于20000000

num = 0 def a(lock): global num for _ in range(1000000): with lock: num += 1 def b(lock): global num for _ in range(1000000): with lock: num += 1 if __name__ == '__main__': lock = threading.Lock() t1=Thread(target=a, args=(lock,)) t1.start() t2=Thread(target=b, args=(lock,)) t2.start() t1.join() t2.join() print(num) #永远会输出20000000

#在之前的代码中永远不可能出现锁在没释放之前重新获得锁，但rlock可以做到，但只能发生在一个线程中，如： num = 0 def a(lock): with lock: print("我是A") b(lock) def b(lock): with lock: print("我是b") if __name__ == '__main__': lock = threading.Lock() t1 = Thread(target=a, args=(lock,)) t1.start() #会发生死锁，因为在第一次还没释放锁后，b就准备上锁，并阻止a释放锁

if __name__ == '__main__': lock = threading.RLock() #只需要改变锁为RLock程序马上恢复 t1 = Thread(target=a, args=(lock,)) t1.start()

#这个程序我们模拟甲乙对话 Jlist = ["在吗", "干啥呢", "去玩儿不", "好吧"] Ylist = ["在呀", "玩儿手机", "不去"] def J(list): for i in list: print(i) time.sleep(0.1) def Y(list): for i in list: print(i) time.sleep(0.1) if __name__ == '__main__': t1 = Thread(target=J, args=(Jlist,)) t1.start() t1.join() t2 = Thread(target=Y, args=(Ylist,)) t2.start() t2.join() #上面的程序输出后发现效果就是咱们想要的，但是我们每次输出后都要等待0.1秒，也无法正好确定可以拿到时间片的最短时间值，并且不能保证每次正好都是另一个线程执行。因此，我们用以下方式，完美解决这些问题。

Jlist = ["在吗", "干啥呢", "去玩儿不", "好吧"] Ylist = ["在呀", "玩儿手机", "不去","哦"] def J(cond, list): for i in list: with cond: print(i) cond.notify() cond.wait() def Y(cond, list): for i in list: with cond: cond.wait() print(i) cond.notify() if __name__ == '__main__': cond = threading.Condition() t1 = Thread(target=J, args=(cond, Jlist)) t2 = Thread(target=Y, args=(cond, Ylist)) t2.start() t1.start() #一定保证t1启动在t2之后,因为notify发送的信号要被t2接受到，如果t1先启动，会发生阻塞。

class B(threading.Thread): def __init__(self, name): super().__init__() self.name = name def run(self): time.sleep(1) print(self.name) class A(threading.Thread): def __init__(self): super().__init__() def run(self): for i in range(100): b = B(i) b.start() if __name__ == '__main__': a = A() a.start() #执行后发现不断在输出

class B(threading.Thread): def __init__(self, name, sem): super().__init__() self.name = name self.sem = sem def run(self): time.sleep(1) print(self.name) sem.release() class A(threading.Thread): def __init__(self, sem): super().__init__() self.sem = sem def run(self): for i in range(100): self.sem.acquire() b = B(i, self.sem) b.start() if __name__ == '__main__': sem = threading.Semaphore(value=3) a = A(sem) a.start() #通过执行上面的代码，我们发现一次只能输出三个数字，sem控制访问并发量