python之迭代锁与信号量

Posted 2021-01-31 晴晴小可爱的小弟

如果现在需要在多处加锁大于等于2的时候 因为计算机比较笨，当他锁上一把锁的时候又所理一把锁，等他来开锁的时候他不知道用哪把钥匙来开锁，

所以这个时候我们需要把把平常的锁变为迭代锁

eg：

import threading
import time

local = threading.RLock()
# 迭代加锁首先生成实例
def run(name):
    global num
    local.acquire()  # 上锁
    num += 1
    run2(num)
    local.release()  # 解锁
    print(threading.active_count())

def run2(num):
    local.acquire()
    num += 1
    run3(num)
    local.release()

def run3(num):
    local.acquire()
    num+=1
    local.release()

num = 0
py_res = []
for i in range(50):
    t = threading.Thread(target=run,args=(‘t%s‘%i,))
    py_res.append(t)
    t.start()


for i in py_res:
    i.join()

print("num : %s"% num)

型号量可以控制同线程的个数，和锁的用法一样

import threading
import time
senm = threading.BoundedSemaphore(5)
def run(num):
    senm.acquire()  # 上锁
    time.sleep(2)
    print(‘run the thread %s‘ % num)
    senm.release()  # 解锁

for i in range(50):
    t = threading.Thread(target=run, args=(‘%s‘ % i,))
    t.start()
    print(threading.active_count())

while threading.active_count() != 1:
    pass
else:
    print(‘is done‘)