#!-*-coding:utf-8-*-
#加用户态的锁，与全局解释器锁不一样(GIL)
import threading, time
def run(n):
    lock.acquire() #获取用户态锁 也叫互斥锁Mutex
    global num #操作
    #time.sleep(0.1) #加了sleep之后程序变串行的了 一般不要加
    num +=1
    lock.release() #释放用户态锁

lock=threading.Lock()
num=0
t_objs = []
for i in range(1000):
    t = threading.Thread(target=run, args=("t %s" % i,))
    t.start()
    t_objs.append(t)  # 把每个线程实例都加进来 不阻塞后面线程的启动
for t in t_objs: #取列表里的每个线程
    t.join() #等待并行的每个线程全都执行完毕 在往下走

print("----all threads has finished...",threading.current_thread(),threading.active_count())
print("num",num)

import threading, timedef run1():     print("grab the first part data")     lock.acquire()     global num     num += 1     lock.release()     return num

def run2():     print("grab the second part data")     lock.acquire()     global num2     num2 += 1     lock.release()     return num2

def run3():     lock.acquire()     res = run1()     print('--------between run1 and run2-----')     res2 = run2()     lock.release()     print(res, res2)

 num, num2 = 0, 0lock = threading.RLock() #递归锁for i in range(1):     t = threading.Thread(target=run3)     t.start()

while threading.active_count() != 1:     print(threading.active_count())else:     print('----all threads done---')     print(num, num2)

import threading,timedef run(n):     semaphore.acquire() #信号量 相当于若干把锁 同时开启5个线程     time.sleep(1)     print("run the thread:%s\n"%n)     semaphore.release() #同时释放5把锁

if __name__=="__main__":     semaphore=threading.BoundedSemaphore(5) #最多允许5个线程同时运行 相当于有5把锁     for i in range(20):         t=threading.Thread(target=run,args=(i,))         t.start()

while threading.active_count() !=1:     pass else:     print("---all threads done----")