Question

我想运行一个项目（project.py），它有两个调度元素，比如我在堆栈溢出时找到的答案：Periodically execute function in thread in real time, every N seconds

一个功能是每x秒运行另一个y秒其中0.03 < x < 0.1和2 < y < 10。

function_x(t)调用暂停（SIGSTOP）project.py

的外部程序（sync.py）

问题0：调用do_every后，程序继续执行。这是正确的假设吗？

问题1：如果在project.py上调用SIGCONT之前sync.py需要2秒才能执行，function_x的计时器是否会被搞砸了＆＃39;？

问题3：如果function_y正在通过套接字发送数据，function_x调用SIGSTOP会导致执行错误吗？时间会暂停吗？计时器如何计算时间？

前面提到的答案：

import threading;

def do_every (interval, worker_func, iterations = 0):
  if iterations != 1:
    threading.Timer (
      interval,
      do_every, [interval, worker_func, 0 if iterations == 0 else iterations-1]
    ).start ();

  worker_func ();

def print_hw ():
  print "hello world";

def print_so ():
  print "stackoverflow"


# call print_so every second, 5 times total
do_every (1, print_so, 5);

# call print_hw two times per second, forever
do_every (0.5, print_hw);

Answer 1

0) After calling do_every - it runs the worker_func once and then continues. yes. So your output will be:

stackoverflow
After first do_every
hello world
After second do_every
hello world
stackoverflow
....

1) I didn't really know if the timer would be messed up - so I ran a test using htop where I can send SIGCONT and SIGSTOP to any process - so, I ran do_every (2, print_so, 10); and after 5-6 "stackoverflow"s I stopped it. After a minute on starting it - it spewed out 4 more stackoverflows. So, no - the timer is not messed up.

3) It will not be messed up - what will happen is that if the socket has a timeout - and you pause for more than the timeout, the other side of the socket will assume your code is dead and will give a timeout exception. But if it's for less than the timeout, the socket will assume it was a temporary disconnection, and will continue streaming the data after that.

定期执行function / SIGSTOP

1 个答案: