📅  最后修改于: 2023-12-03 15:12:39.298000             🧑  作者: Mango
这道问题是关于多线程和同步的,需要合理地运用锁来解决线程间资源共享的问题。
有三个线程A、B、C,它们分别打印字母A、B、C,要求这三个线程一起运行,打印n次,输出格式为ABCABC....
这道问题的难点在于如何实现三个线程的协调和同步。有多种方式可以实现,例如使用Java中的wait和notify方法、使用信号量进行同步等。这里我们介绍一种比较简单易懂的方式:使用锁和条件变量。
首先定义一个锁和两个条件变量。锁用来保证线程之间的互斥,而条件变量则用来实现线程的等待和唤醒。
import threading
lock = threading.Lock()
condition_b = threading.Condition(lock)
condition_c = threading.Condition(lock)
接着定义三个线程的函数,每个线程里面有一个循环,循环n次,每次先获取锁,然后判断自己是否可以打印。如果不能打印,就释放锁,等待条件变量唤醒;如果能打印,就打印相应的字符,然后唤醒下一个线程,并释放锁。
def thread_a(n):
for i in range(n):
with lock:
print('A', end='')
condition_b.notify()
condition_c.notify()
if i != n - 1:
condition_c.wait()
else:
print()
def thread_b(n):
for i in range(n):
with lock:
condition_b.wait()
print('B', end='')
condition_c.notify()
if i != n - 1:
condition_c.wait()
else:
print()
def thread_c(n):
for i in range(n):
with lock:
condition_c.wait()
print('C', end='')
condition_b.notify()
if i != n - 1:
condition_c.wait()
else:
print()
最后在主函数中创建三个线程并启动它们:
if __name__ == '__main__':
n = 10
t1 = threading.Thread(target=thread_a, args=(n,))
t2 = threading.Thread(target=thread_b, args=(n,))
t3 = threading.Thread(target=thread_c, args=(n,))
t1.start()
t2.start()
t3.start()
t1.join()
t2.join()
t3.join()
import threading
lock = threading.Lock()
condition_b = threading.Condition(lock)
condition_c = threading.Condition(lock)
def thread_a(n):
for i in range(n):
with lock:
print('A', end='')
condition_b.notify()
condition_c.notify()
if i != n - 1:
condition_c.wait()
else:
print()
def thread_b(n):
for i in range(n):
with lock:
condition_b.wait()
print('B', end='')
condition_c.notify()
if i != n - 1:
condition_c.wait()
else:
print()
def thread_c(n):
for i in range(n):
with lock:
condition_c.wait()
print('C', end='')
condition_b.notify()
if i != n - 1:
condition_c.wait()
else:
print()
if __name__ == '__main__':
n = 10
t1 = threading.Thread(target=thread_a, args=(n,))
t2 = threading.Thread(target=thread_b, args=(n,))
t3 = threading.Thread(target=thread_c, args=(n,))
t1.start()
t2.start()
t3.start()
t1.join()
t2.join()
t3.join()
以上就是本题的解法和代码,希望对大家有所帮助。