How does polling a file for changes work?

Question

The problem

I expected the script below to print at most one event and then stop (it's written only to illustrate the problem).

#!/usr/bin/env python

from select import poll, POLLIN

filename = "test.tmp"

# make sure file exists
open(filename, "a").close()

file = open(filename, "r+")

p = poll()
p.register(file.fileno(), POLLIN)

while True:
    events = p.poll(100)
    for e in events:
        print e
        # Read data, so that the event goes away?
        file.read()

However, it prints about 70000 events per second. Why?

Background

I've written a class that uses the pyudev.Monitor class internally. Amongst other things, it polls the fileno supplied by the fileno() method for changes using a poll object.

Now I'm trying to write an unit test for my class (I realize I'm supposed to write the unit test first, so no need to point it out), and therefore I need to write my own fileno() method for my mock pyudev.Monitor object, and I need to control it so that I can trigger the poll object to report an event. As the above code demonstrates, I can't make it stop reporting seemingly non-existent events!

I can find no acknowledge_event() or similar in the poll class to make the event go away (I suspect there's just one event that's somehow stuck), searching google and this site has yielded nothing. I'm using python 2.6.6 on Ubuntu 10.10.

Capsular

Answer 1

You'll have better luck using pipes rather than files. Try this instead:

#!/usr/bin/env python
import os
from   select import poll, POLLIN

r_fd, w_fd = os.pipe()

p = poll()
p.register(r_fd, POLLIN)

os.write(w_fd, 'X') # Put something in the pipe so p.poll() will return

while True:
    events = p.poll(100)
    for e in events:
        print e
        os.read(r_fd, 1)

This will print out the single event you're looking for. To trigger the poll event, all you have to do is write a byte to the writeable file descriptor.

Olsson