How to let the child process live when parent process exited?

Question

I want to use multiprocessing module to complete this.

when I do this, like:

    $ python my_process.py

I start a parent process, and then let the parent process spawn a child process,

then i want that the parent process exits itself, but the child process continues to work.

Allow me write a WRONG code to explain myself:

from multiprocessing import Process

def f(x):
    with open('out.dat', 'w') as f:
        f.write(x)

if __name__ == '__main__':
    p = Process(target=f, args=('bbb',))
    p.daemon = True    # This is key, set the daemon, then parent exits itself
    p.start()

    #p.join()    # This is WRONG code, just want to exlain what I mean.
    # the child processes will be killed, when father exit

So, how do i start a process that will not be killed when the parent process finishes?

---

20140714

Hi, you guys

My friend just told me a solution...

I just think...

Anyway, just let u see:

import os
os.system('python your_app.py&')    # SEE!? the & !!

this does work!!

Answer 1

A trick: call os._exit to make parent process exit, in this way daemonic child processes will not be killed.

But there are some other side affects, described in the doc:

Exit the process with status n, without calling cleanup handlers, 
flushing stdio buffers, etc.

If you do not care about this, you can use it.