Question

我正在使用Popen启动子进程，我希望它能完成退出。但是，这个过程不仅没有退出，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，打印：

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

这个脚本不仅可以验证孩子在完成后仍然活着，但是我可以在打印的进程ID上使用ps并看到它仍然存在。奇怪的是，ps将进程名称列为(Python)（注意括号）。

Answer 1

您需要拨打process.wait()或使用signal.signal(signal.SIGCHLD, signal.SIG_IGN)一次表示您不打算等待任何孩子。前者是便携式的;后者仅适用于Unix（但POSIX-standard）。如果你不做这些事情，那么在Unix上，这个过程就像僵尸一样，以及Windows has a similar behavior if you keep the process handle open。

为什么Popen儿童过程不会死？

1 个答案: