Ensure a single instance of an application in Linux

Question

I'm working on a GUI application in WxPython, and I am not sure how I can ensure that only one copy of my application is running at any given time on the machine. Due to the nature of the application, running more than once doesn't make any sense, and will fail quickly. Under Win32, I can simply make a named mutex and check that at startup. Unfortunately, I don't know of any facilities in Linux that can do this.

I'm looking for something that will automatically be released should the application crash unexpectedly. I don't want to have to burden my users with having to manually delete lock files because I crashed.

Answer 1

The Right Thing is advisory locking using flock(LOCK_EX); in Python, this is found in the fcntl module.

Unlike pidfiles, these locks are always automatically released when your process dies for any reason, have no race conditions exist relating to file deletion (as the file doesn't need to be deleted to release the lock), and there's no chance of a different process inheriting the PID and thus appearing to validate a stale lock.

If you want unclean shutdown detection, you can write a marker (such as your PID, for traditionalists) into the file after grabbing the lock, and then truncate the file to 0-byte status before a clean shutdown (while the lock is being held); thus, if the lock is not held and the file is non-empty, an unclean shutdown is indicated.

Answer 2

Complete locking solution using the fcntl module:

import fcntl
pid_file = 'program.pid'
fp = open(pid_file, 'w')
try:
    fcntl.lockf(fp, fcntl.LOCK_EX | fcntl.LOCK_NB)
except IOError:
    # another instance is running
    sys.exit(1)

Answer 3

There are several common techniques including using semaphores. The one I see used most often is to create a "pid lock file" on startup that contains the pid of the running process. If the file already exists when the program starts up, open it up and grab the pid inside, check to see if a process with that pid is running, if it is check the cmdline value in /proc/pid to see if it is an instance of your program, if it is then quit, otherwise overwrite the file with your pid. The usual name for the pid file is application_name.pid.

Answer 4

wxWidgets offers a wxSingleInstanceChecker class for this purpose: wxPython doc, or wxWidgets doc. The wxWidgets doc has sample code in C++, but the python equivalent should be something like this (untested):

  name = "MyApp-%s" % wx.GetUserId()
  checker = wx.SingleInstanceChecker(name)
  if checker.IsAnotherRunning():
      return False

Answer 5

This builds upon the answer by user zgoda. It mainly addresses a tricky concern having to do with write access to the lock file. In particular, if the lock file was first created by root, another user foo can then no successfully longer attempt to rewrite this file due to an absence of write permissions for user foo. The obvious solution seems to be to create the file with write permissions for everyone. This solution also builds upon a different answer by me, having to do creating a file with such custom permissions. This concern is important in the real world where your program may be run by any user including root.

import fcntl, os, stat, tempfile

app_name = 'myapp'  # <-- Customize this value

# Establish lock file settings
lf_name = '.{}.lock'.format(app_name)
lf_path = os.path.join(tempfile.gettempdir(), lf_name)
lf_flags = os.O_WRONLY | os.O_CREAT
lf_mode = stat.S_IWUSR | stat.S_IWGRP | stat.S_IWOTH  # This is 0o222, i.e. 146

# Create lock file
# Regarding umask, see https://stackoverflow.com/a/15015748/832230
umask_original = os.umask(0)
try:
    lf_fd = os.open(lf_path, lf_flags, lf_mode)
finally:
    os.umask(umask_original)

# Try locking the file
try:
    fcntl.lockf(lf_fd, fcntl.LOCK_EX | fcntl.LOCK_NB)
except IOError:
    msg = ('Error: {} may already be running. Only one instance of it '
           'can run at a time.'
           ).format('appname')
    exit(msg)

A limitation of the above code is that if the lock file already existed with unexpected permissions, those permissions will not be corrected.

I would've liked to use /var/run/<appname>/ as the directory for the lock file, but creating this directory requires root permissions. You can make your own decision for which directory to use.

Note that there is no need to open a file handle to the lock file.

Answer 6

Here's the TCP port-based solution:

# Use a listening socket as a mutex against multiple invocations
import socket
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind(('127.0.0.1', 5080))
s.listen(1)

Answer 7

The set of functions defined in semaphore.h -- sem_open(), sem_trywait(), etc -- are the POSIX equivalent, I believe.

Answer 8

Look for a python module that interfaces to SYSV semaphores on unix. The semaphores have a SEM_UNDO flag which will cause the resources held by the a process to be released if the process crashes.

Otherwise as Bernard suggested, you can use

import os
os.getpid()

And write it to /var/run/application_name.pid. When the process starts, it should check if the pid in /var/run/application_name.pid is listed in the ps table and quit if it is, otherwise write its own pid into /var/run/application_name.pid. In the following var_run_pid is the pid you read from /var/run/application_name.pid

cmd = "ps -p %s -o comm=" % var_run_pid
app_name = os.popen(cmd).read().strip()
if len(app_name) > 0:
    Already running

Answer 9

If you create a lock file and put the pid in it, you can check your process id against it and tell if you crashed, no?

I haven't done this personally, so take with appropriate amounts of salt. :p

Answer 10

Can you use the 'pidof' utility? If your app is running, pidof will write the Process ID of your app to stdout. If not, it will print a newline (LF) and return an error code.

Example (from bash, for simplicity):

linux# pidof myapp
8947
linux# pidof nonexistent_app

linux#

Answer 11

By far the most common method is to drop a file into /var/run/ called [application].pid which contains only the PID of the running process, or parent process. As an alternative, you can create a named pipe in the same directory to be able to send messages to the active process, e.g. to open a new file.

Answer 12

I've made a basic framework for running these kinds of applications when you want to be able to pass the command line arguments of subsequent attempted instances to the first one. An instance will start listening on a predefined port if it does not find an instance already listening there. If an instance already exists, it sends its command line arguments over the socket and exits.

code w/ explanation