python - 子进程超时失败

Question

我想在子进程上使用超时

 from subprocess32 import check_output
 output = check_output("sleep 30", shell=True, timeout=1)

不幸的是，虽然这会引发超时错误，但它会在 30 秒后发生。 check_output好像不能打断shell命令。

我可以在 Python 方面做些什么来阻止这种情况？ 我怀疑 subprocess32 未能终止超时进程。

Answer 1

check_output() with timeout is essentially :

with Popen(*popenargs, stdout=PIPE, **kwargs) as process:
    try:
        output, unused_err = process.communicate(inputdata, timeout=timeout)
    except TimeoutExpired:
        process.kill()
        output, unused_err = process.communicate()
        raise TimeoutExpired(process.args, timeout, output=output)

有两个问题:

• [第二个] .communicate() 可能会等待后代进程，而不仅仅是直接子进程，请参阅 Python subprocess .check_call vs .check_output
• process.kill() 可能不会杀死整个进程树，参见 How to terminate a python subprocess launched with shell=True

它导致您观察到的行为:TimeoutExpired 发生在一秒钟内，shell 被杀死，但是 check_output() 仅在孙子之后的 30 秒内返回sleep 进程退出。

要解决这些问题，请终止整个进程树(属于同一组的所有子进程):

#!/usr/bin/env python3
import os
import signal
from subprocess import Popen, PIPE, TimeoutExpired
from time import monotonic as timer

start = timer()
with Popen('sleep 30', shell=True, stdout=PIPE, preexec_fn=os.setsid) as process:
    try:
        output = process.communicate(timeout=1)[0]
    except TimeoutExpired:
        os.killpg(process.pid, signal.SIGINT) # send signal to the process group
        output = process.communicate()[0]
print('Elapsed seconds: {:.2f}'.format(timer() - start))

输出

Elapsed seconds: 1.00