线程信号量和进程信号量相似

# 线程信号量
import time
from threading import Semaphore
from threading import Thread


def test(n, se):
    se.acquire()
    print(pow(n, 2))
    time.sleep(1)
    se.release()


se = Semaphore(4)
for i in range(10):
    t = Thread(target=test, args=(i, se))
    t.start()

 

相关文章:

  • 2021-12-22
  • 2021-06-22
  • 2021-07-30
  • 2022-12-23
  • 2021-08-19
  • 2022-02-27
  • 2021-08-21
  • 2021-10-28
猜你喜欢
  • 2022-01-24
  • 2021-10-11
  • 2022-12-23
  • 2021-10-11
  • 2022-01-15
  • 2021-09-07
  • 2022-12-23
相关资源
相似解决方案