How do I abort reading a user's input in Python 3.x?

Question

I have a thread that monitors user input which looks like this:

def keyboard_monitor(temp): #temp is sys.stdin
global flag_exit

while True:
    keyin = temp.readline().rstrip().lower()
    if keyin == "exit":
        flag_exit = True
        print("Stopping...")

    if flag_exit == True:
        break

If I type exit the flag is properly set and all the other threads terminate. If another one of my threads sets the flag, this thread refuses to finish because it's hanging on the user input. I have to input something to get it to finish. How do I change this code so that the program finishes when the flag is set externally?

you could try `temp.close()`. dont know what effect that might have. — meuh, Sep 11 '15 at 16:00
Investigate Alex Martelli's [Post](http://stackoverflow.com/a/2933423/298607) and use `msvcrt.kbhit()` — dawg, Sep 11 '15 at 17:22
If you want x platform, maybe use [Tkinter](https://en.wikipedia.org/wiki/Tkinter) — dawg, Sep 11 '15 at 17:23

score 0 · Answer 1 · edited Sep 12 '15 at 09:23

0

Its hard to tell exactly what is going wrong without more of your code, but as an easy solution you could rather exit() which is a python built in. This should reliably terminate the application, also sys.exit()

From wim's comment:

You can use the atexit module to register clean up handlers

import atexit

def cleanup():
   pass
   # TODO cleanup

atexit.register(cleanup)

edited Sep 12 '15 at 09:23

wim

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answered Sep 11 '15 at 16:54

Ciaran Liedeman

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When giving an answer with a link, it is usually preferred to copy the critical parts of the solution directly to the answer – wnnmaw Sep 11 '15 at 17:28
Thanks, I'll remember to do that – Ciaran Liedeman Sep 11 '15 at 17:29
I have to do some house keeping (closing file handles etc.) so I can't just call exit. – BB ON Sep 11 '15 at 17:38
Do you know about [`atexit`](https://docs.python.org/2/library/atexit.html) module? – wim Sep 11 '15 at 17:43

How do I abort reading a user's input in Python 3.x?

1 Answers1