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.
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.)
#include <afx.h>
You need a variable of type HANDLE with the same lifetime as your program. Perhaps as a member of your CWinApp object.
HANDLE mutex;
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.
mutex = CreateMutex( NULL, TRUE, "MyApp" ); if ( GetLastError() == ERROR_ALREADY_EXISTS ) { // There's another instance running. What do you do? }
Finally, near the end of your program, you need to close the mutex.
CloseHandle( mutex );
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>
Python
Linux (including cygwin) and Mac OSX Leopard
Must be run from an application, not the interpreter.
import __main__, os def isOnlyInstance(): # Determine if there are more than the current instance of the application # running at the current time. If not, return 0. If so, return 1. if not os.system('(( $(ps -ef | grep python | grep \'[' + \ __main__.__file__[0] + ']' + __main__.__file__[1:len(__main__.__file__)] \ + '\' | wc -l) > 1 ))'): return 0 else: return 1
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.
Tcl
<lang Tcl>set APP_SPECIFIC_PORT_NUMBER 12345 try {
socket -server closePort -myaddr localhost $APP_SPECIFIC_PORT_NUMBER proc closePort {chan add port} {close $chan}
} trap {POSIX EADDRINUSE} {} {
# Generate a nice error message puts stderr "Application $::argv0 already running?" exit
}</lang>
Visual Basic
Dim onlyInstance as Boolean onlyInstance = not App.PrevInstance