Make sure only a single instance of a program is running

Is there a Pythonic way to have only one instance of a program running?

The only reasonable solution I've come up with is trying to run it as a server on some port, then second program trying to bind to same port - fails. But it's not really a great idea, maybe there's something more lightweight than this?

(Take into consideration that program is expected to fail sometimes, i.e. segfault - so things like "lock file" won't work)

Solution 1:

The following code should do the job, it is cross-platform and runs on Python 2.4-3.2. I tested it on Windows, OS X and Linux.

from tendo import singleton
me = singleton.SingleInstance() # will sys.exit(-1) if other instance is running

The latest code version is available singleton.py. Please file bugs here.

You can install tend using one of the following methods:

• easy_install tendo
• pip install tendo
• manually by getting it from http://pypi.python.org/pypi/tendo

Solution 2:

Simple, cross-platform solution, found in another question by zgoda:

import fcntl
import os
import sys

def instance_already_running(label="default"):
    """
    Detect if an an instance with the label is already running, globally
    at the operating system level.

    Using `os.open` ensures that the file pointer won't be closed
    by Python's garbage collector after the function's scope is exited.

    The lock will be released when the program exits, or could be
    released if the file pointer were closed.
    """

    lock_file_pointer = os.open(f"/tmp/instance_{label}.lock", os.O_WRONLY)

    try:
        fcntl.lockf(lock_file_pointer, fcntl.LOCK_EX | fcntl.LOCK_NB)
        already_running = False
    except IOError:
        already_running = True

    return already_running

A lot like S.Lott's suggestion, but with the code.

Solution 3:

This code is Linux specific. It uses 'abstract' UNIX domain sockets, but it is simple and won't leave stale lock files around. I prefer it to the solution above because it doesn't require a specially reserved TCP port.

try:
    import socket
    s = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
    ## Create an abstract socket, by prefixing it with null. 
    s.bind( '\0postconnect_gateway_notify_lock') 
except socket.error as e:
    error_code = e.args[0]
    error_string = e.args[1]
    print "Process already running (%d:%s ). Exiting" % ( error_code, error_string) 
    sys.exit (0) 

The unique string postconnect_gateway_notify_lock can be changed to allow multiple programs that need a single instance enforced.

Solution 4:

I don't know if it's pythonic enough, but in the Java world listening on a defined port is a pretty widely used solution, as it works on all major platforms and doesn't have any problems with crashing programs.

Another advantage of listening to a port is that you could send a command to the running instance. For example when the users starts the program a second time, you could send the running instance a command to tell it to open another window (that's what Firefox does, for example. I don't know if they use TCP ports or named pipes or something like that, 'though).