python - 为什么 Popen 子进程不会死亡？

Question

我正在使用 Popen 启动一个子进程，我希望它能够完成退出。然而，进程不仅没有退出，而且向它发送 sigkill 仍然让它活着!下面是一个演示的脚本:

from subprocess import Popen
import os
import time
import signal

command = ["python","--version"]

process = Popen(command)
pid = process.pid

time.sleep(5) #ample time to finish

print pid

print "Sending sigkill"
os.kill(pid,signal.SIGKILL)

try:
    #Kill with signal 0 just checks whether process exists
    os.kill(pid,0)
    print "Process still alive immediately after (not so bad...)!"
except Exception as e:
    print "Succeeded in terminating child quickly!"

time.sleep(20) #Give it ample time to die

#Kill with signal 0 just checks whether process exists
try:
    os.kill(pid,0)
    print "Process still alive! That's bad!"
except Exception as e:
    print "Succeeded in terminating child!"

对我来说，这打印:

77881
Python 2.7.10
Sending sigkill
Process still alive immediately after (not so bad...)!
Process still alive! That's bad!

此脚本不仅可以验证子进程在应该完成后是否仍然存在，而且我可以在打印的进程 ID 上使用 ps 并查看它是否仍然存在。奇怪的是， ps 将进程名称列为 (Python) (注意括号)。

Answer 1

您需要调用 process.wait() ，或使用signal.signal(signal.SIGCHLD, signal.SIG_IGN)一次表明您不打算等待任何 child 。前者是可移植的；后者仅适用于 Unix(但为 POSIX-standard )。如果你不做这些事情，在 Unix 上，进程将像僵尸一样徘徊，而 Windows has a similar behavior if you keep the process handle open .