标签：thread def 线程 变量 reading 解决 print 竞争 random
1线程的创建：
import threading
import time,random
def text1():
while True:
print(1111111)
time.sleep(random.random()*2)
def text2():
while True:
print(2222222)
time.sleep(random.random() * 2)
def main():
# text1()
# text2()
#创建多线程
t1 = threading.Thread(target=text1)
t2 = threading.Thread(target=text2)
t1.start() #执行多线程
t2.start()
if __name__ == "__main__":
main()
2互斥锁：
在多线程之中全局变量是共享的；在执行过程中又可能会发生资源竞争，所以会用到互斥锁：比如
import threading
import time,os,random
num = 0
def text1(agr):
global num
for i in range(agr):
num += 1
print(num)
def text2(agr):
global num
for i in range(agr):
num += 1
print(num)
def main():
t1 = threading.Thread(target=text1,args=(1000000,))
t2 = threading.Thread(target=text2,args=(1000000,))
t1.start()
t2.start()
time.sleep(5)
print(num)
if __name__ == "__main__":
main()
执行结果：如下，而不是我们向看到的2000000
1170362 1302259 1302259
如何解决呢，用到互斥锁：
import threading
import time,os,random
num = 0
def text1(agr,mutex):
global num
for i in range(agr):
mutex.acquire() #上锁
num += 1
mutex.release() #解锁
print(num)
def text2(agr,mutex):
global num
for i in range(agr):
mutex.acquire() #上锁
num += 1
mutex.release() #解锁
print(num)
def main():
mutex = threading.Lock() #创建一个互斥锁
t1 = threading.Thread(target=text1,args=(1000000,mutex))
t2 = threading.Thread(target=text2,args=(1000000,mutex))
t1.start()
t2.start()
time.sleep(5)
print(num)
if __name__ == "__main__":
main()
结果：
1846157 2000000 2000000
标签：thread def 线程 变量 reading 解决 print 竞争 random
原文地址：https://www.cnblogs.com/yan-peng/p/9972606.html
