Determine if only one instance is running
You are encouraged to solve this task according to the task description, using any language you may know.
This task is to determine if there is only one instance of an application running.
AutoHotkey
AutoHotkey has a #SingleInstance command. If you run two scripts that don't have it at the same time, it alerts the user. #SingleInstance FORCE closes the older instance when a newer one is run, and #SingleInstance IGNORE does nothing when you try to open a new instance of an already-running script.
C
Notes: this should work on every POSIX compliant system (you need to link with the POSIX thread pthread library).
It must be noted that if the program is interrupted and the sem_unlink is not executed, the semaphore will be still around preventing another execution of the program. So the program, when failing, must be sure to run the sem_unlink statement (e.g. doing it in a function that is passed to atexit and catching killing signals properly).
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <semaphore.h>
- include <unistd.h>
- include <fcntl.h>
/* unistd for sleep */
int main() {
sem_t *mysem; mysem = sem_open("MyUniqueName", O_CREAT|O_EXCL); if ( mysem == NULL ) { fprintf(stderr, "I am already running!\n"); exit(1); } /* here the real code of the app*/ sleep(20); /* end of the app */ sem_unlink("MyUniqueName"); sem_close(mysem);
}</lang>
C++
Microsoft Windows
This line needs to be near the top of the file (or in stdafx.h, if you use one.) <lang cpp>#include <afx.h></lang>
You need a variable of type HANDLE with the same lifetime as your program. Perhaps as a member of your CWinApp object. <lang cpp>HANDLE mutex;</lang>
At the earliest possible point in your program, you need to initialize it and perform your check. "MyApp" should be a string unique to your application. See here for full details.
<lang cpp>mutex = CreateMutex( NULL, TRUE, "MyApp" ); if ( GetLastError() == ERROR_ALREADY_EXISTS ) {
// There's another instance running. What do you do?
}</lang>
Finally, near the end of your program, you need to close the mutex. <lang cpp>CloseHandle( mutex );</lang>
C#
<lang csharp>using System; using System.Net; using System.Net.Sockets;
class Program {
static void Main(string[] args) { try { TcpListener server = new TcpListener(IPAddress.Any, 12345); server.Start(); } catch (SocketException e) { if (e.SocketErrorCode == SocketError.AddressAlreadyInUse) { Console.Error.WriteLine("Already running."); } } }
}</lang>
Java
<lang java>import java.io.IOExeception; import java.net.InetAddress; import java.net.ServerSocket; import java.net.UnknownHostException;
public class SingletonApp { private static final int PORT = 12345; // random large port number private static ServerSocket s;
// static initializer { try { s = new ServerSocket(PORT, 10, InetAddress.getLocalHost()); } catch (UnknownHostException e) { // shouldn't happen for localhost } catch (IOException e) { // port taken, so app is already running System.exit(0); } } // main() and rest of application... }</lang>
Liberty BASIC
<lang lb>'Create a Mutex to prevent more than one instance from being open at a single time. CallDLL #kernel32, "CreateMutexA", 0 as Long, 1 as Long, "Global\My Program" as ptr, mutex as ulong CallDLL #kernel32, "GetLastError", LastError as Long
if LastError = 183 then 'Error returned when a Mutex already exists
'Close the handle if the mutex already exists calldll #kernel32, "CloseHandle", mutex as ulong, ret as ulong notice "An instance of My Program is currently running!" end
end if
'Release the Mutex/ Close the handle prior to ending the program 'Comment out these lines to allow the program to remain active to test for the mutex's presence calldll #kernel32, "ReleaseMutex", mutex as ulong, ret as ulong calldll #kernel32, "CloseHandle", mutex as ulong, ret as ulong end</lang>
OCaml
Replicates the C example, with the library ocaml-sem. <lang ocaml>open Sem
let () =
let oflags = [Unix.O_CREAT; Unix.O_EXCL] in let sem = sem_open "MyUniqueName" ~oflags () in (* here the real code of the app *) Unix.sleep 20; (* end of the app *) sem_unlink "MyUniqueName"; sem_close sem</lang>
The standard library of OCaml also provides a Mutex module.
Oz
<lang oz>functor import Application Open System define
fun {IsAlreadyRunning} try
S = {New Open.socket init}
in
{S bind(takePort:12345)} false
catch system(os(os "bind" ...) ...) then
true
end end
if {IsAlreadyRunning} then {System.showInfo "Exiting because already running."} {Application.exit 1} end {System.showInfo "Press enter to exit."} {{New Open.file init(name:stdin)} read(list:_ size:1)} {Application.exit 0}
end</lang>
Perl
The INIT block is runned just before the Perl runtime begins execution. See perlmod
Then it tries to get a lock to its own file, from where the script was called. <lang perl>use Fcntl ':flock';
INIT { die "Not able to open $0\n" unless (open ME, $0); die "I'm already running !!\n" unless(flock ME, LOCK_EX|LOCK_NB); }
sleep 60; # then your code goes here</lang>
PicoLisp
Calling 'killall'
One possibility is to send a zero-signal with 'killall', and check the return value. This is useful if each application is started by a hash-bang script (the first line is e.g. "#!/usr/bin/picolisp /usr/lib/picolisp/lib.l"). In that way, each application has its onw name which can be passed to 'killall'.
$ cat myScript #!/usr/bin/picolisp /usr/lib/picolisp/lib.l (wait 120000) (bye)
$ ./myScript & # Start in the background [1] 26438
<lang PicoLisp>$ ./dbg
- (call "killall" "-0" "-q" "myScript")
-> T</lang>
Using a mutex
Another possibility is to 'acquire' a mutex on program start, and never release it. <lang PicoLisp>: (acquire "running1") -> 30817 # A successful call returns the PID</lang> A second application trying to acquire the same mutex would receive 'NIL'
PureBasic
<lang PureBasic>#MyApp="MyLittleApp" Mutex=CreateMutex_(0,1,#MyApp) If GetLastError_()=#ERROR_ALREADY_EXISTS
MessageRequester(#MyApp,"One instance is already started.") End
EndIf
- Main code executes here
ReleaseMutex_(Mutex) End</lang>
Python
Linux (including cygwin) and Mac OSX Leopard
Must be run from an application, not the interpreter.
<lang python>import __main__, os
def isOnlyInstance():
# Determine if there are more than the current instance of the application # running at the current time. return os.system("(( $(ps -ef | grep python | grep '[" + __main__.__file__[0] + "]" + __main__.__file__[1:] + "' | wc -l) > 1 ))") != 0</lang>
This is not a solution - one can run the same app by copying the code to another location. A solution may be a lock file or lock directory created by the first instance and hold while the first instance is running.
Ruby
Uses file locking on the program file <lang ruby>def main
puts "first instance" sleep 20 puts :done
end
if $0 == __FILE__
if File.new(__FILE__).flock(File::LOCK_EX | File::LOCK_NB) main else raise "another instance of this program is running" end
end
__END__</lang>
Tcl
<lang Tcl>package require Tcl 8.6 try {
# Pick a port number based on the name of the main script executing socket -server {apply {{chan args} {close $chan}}} -myaddr localhost \ [expr {1024 + [zlib crc32 [file normalize $::argv0]] % 30000}]
} trap {POSIX EADDRINUSE} {} {
# Generate a nice error message puts stderr "Application $::argv0 already running?" exit 1
}</lang>
Visual Basic
<lang vb>Dim onlyInstance as Boolean onlyInstance = not App.PrevInstance</lang>