semaphore是一个内置的计数器
计数器不能小于0,当计数器为0时,acquire()将阻塞线程直到其他线程调用release()。
来看下面的代码:
import time
import threading
def foo():
time.sleep(2) #程序休息2秒
print("ok",time.ctime())
for i in range(20):
t1=threading.Thread(target=foo,args=()) #实例化一个线程
t1.start() #启动线程
执行结果:
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
ok Tue Jul 18 20:05:58 2017
可以看到,程序会在很短的时间内生成20个线程来打印一句话。
如果在主机执行IO密集型任务的时候再执行这种类型的程序时,计算机就有很大可能会宕机。
这时候就可以为这段程序添加一个计数器功能,来限制一个时间点内的线程数量。
代码如下:
import time
import threading
s1=threading.Semaphore(5) #添加一个计数器
def foo():
s1.acquire() #计数器获得锁
time.sleep(2) #程序休眠2秒
print("ok",time.ctime())
s1.release() #计数器释放锁
for i in range(20):
t1=threading.Thread(target=foo,args=()) #创建线程
t1.start() #启动线程
执行结果:
ok Tue Jul 18 20:04:38 2017
ok Tue Jul 18 20:04:38 2017
ok Tue Jul 18 20:04:38 2017
ok Tue Jul 18 20:04:38 2017
ok Tue Jul 18 20:04:38 2017
ok Tue Jul 18 20:04:40 2017
ok Tue Jul 18 20:04:40 2017
ok Tue Jul 18 20:04:40 2017
ok Tue Jul 18 20:04:40 2017
ok Tue Jul 18 20:04:40 2017
ok Tue Jul 18 20:04:42 2017
ok Tue Jul 18 20:04:42 2017
ok Tue Jul 18 20:04:42 2017
ok Tue Jul 18 20:04:42 2017
ok Tue Jul 18 20:04:42 2017
ok Tue Jul 18 20:04:44 2017
ok Tue Jul 18 20:04:44 2017
ok Tue Jul 18 20:04:44 2017
ok Tue Jul 18 20:04:44 2017
ok Tue Jul 18 20:04:44 2017