Under win32 I could create a mutex to check if there was already an instance of an application running. What's the recommended way to do the same check under linux/bsd? Should I create a mutex with <pthreads.h>?
Under win32 I could create a mutex to check if there was already an instance of an application running. What's the recommended way to do the same check under linux/bsd? Should I create a mutex with <pthreads.h>?
Here's one technique
http://lists.wxwidgets.org/archive/w.../msg20443.html
It would work on windows as well (as a bonus)
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Catch with that technique is, if another program uses the same socket, that you may get unintended interactions between it and your program. Of course, the odds of that are low. It is reasonably unlikely that two programs will use the same socket. Then the problem to be considered is that of how both programs respond to data they receive over the socket (eg you send data down that happens to trigger a response you didn't intend from the other program).
I also have the same CreateMutex problem. Here's a possible solution using shared memory that I haven't implemented yet.
A scenario would be that the first program (progam A) would initially create the shared memory and post a message to the shared memory indicating that it is running. Now the second program (program b) tries to create the shared memory but it fails because of IPC_EXCL. Program B will now read the shared memory to determine if another instance is running. If another instance is running, then program B will terminate. Otherwise, program B will post a message to shared memory indicating that it is running and will start executing since program A on termination, has already posted a message to shared memory indicating that it is no longer running .Code:key = 5678; if ((shmid = shmget(key, 27, IPC_CREAT | IPC_EXCL | 0666)) < 0) { /* Create failed */ /* Read shared memory to determine if another app is running */ /* If the app is running, then exit else fire up */ }
Thus, a second instance cannot execute until shared memory indicates that it is OK to run.
Anybody have any suggestions, comments etc?
Last edited by BobS0327; 12-09-2006 at 07:45 PM.
I was always under the impression that a mutex was a binary semaphore. So either having value 0 or 1. I donno where this shared memory stuff is commin from. If this is the case sem.h would be what you wanna look at
In a Windows environment, we would use the following code to prevent a second instance of an app from executing while the first instance is still running. In other words, the end user cannot have the same app running two or more times simultaneously. I'm trying to determine the Linux equivalent of the following code:
Code:// Generated this GUID using Guidgen to eliminate multiple instances of fidoh. #define UNIQUE "fidoh-{0xc1bde1e, 0xffc1, 0x4cde, 0x9f, 0x45, 0xfe, 0x6e, 0x5f, 0x4d, 0xc, 0x24)" HANDLE hMutex; hMutex = CreateMutex(NULL,TRUE, UNIQUE); // Check to determine if we're already running if(GetLastError() == ERROR_ALREADY_EXISTS) { MessageBox(NULL, " Application is already running!!! ",MESSAGE_BOX_TITLE,MB_OK | MB_ICONEXCLAMATION); return FALSE; }
so create a binary semaphore with a prechosen ID, so when the program starts it "attaches" (for lack of a better word) to that semaphore. if the semaphore value is 1 (which it is initialized as), the program decreases the value to 0. if its not 1, the program can go no further so you know another instance must be open, and has already decremented the value of the semaphore.
::edit - and also when the program is done executing it has to increase the semaphore value back to 1
linux example of only allowing 1 instance of a program:
initialization program
now the main programCode:#include <stdio.h> #include <stdlib.h> #include <sys/sem.h> #include <sys/types.h> #include <sys/ipc.h> #define SEMID 313131 /* remember this value, you'll need it later */ union semun { int val; struct semid_ds *buf; unsigned short *array; }; int main(void) { int sem_set_id, ret_val; union semun sem_val; struct sembuf sem_op; sem_val.val = 1; /* initialization value */ /* create the semaphore with the specified ID */ sem_set_id = semget(SEMID, 1, IPC_CREAT | 0666); if (sem_set_id == -1) { perror("semget"); exit(1); } /* initialize semaphore */ ret_val = semctl(sem_set_id, 0, SETVAL, sem_val); if (ret_val == -1) { perror("semctl"); exit(1); } return 0; }
output:Code:#include <sys/types.h> #include <sys/ipc.h> #include <unistd.h> #define SEMID 313131 /* notice same ID */ int main(void) { int sem_set_id, ret_val; struct sembuf sem_op; char ch; /* attach to semaphore */ sem_set_id = semget(SEMID, 1, 0666); if (sem_set_id == -1) { perror("semget"); exit(1); } /* check if semaphore is < 1 */ sem_op.sem_num = 0; sem_op.sem_op = -1; sem_op.sem_flg = IPC_NOWAIT; /* don't wait if in use */ if (semop(sem_set_id, &sem_op, 1) == -1) { fprintf(stderr, "Sorry, only 1 instance of program allowed\n"); exit(1); } /* we're here there must be no other instance running */ printf("No other instance running\n"); while (1) { ch = getchar(); if (ch == 'x') break; } /* increase value back to 1 */ sem_op.sem_num = 0; sem_op.sem_op = 1; sem_op.sem_flg = 0; semop(sem_set_id, &sem_op, 1); return 0; }
./init_program
./main
No other instance running
goto another terminal
./main
Sorry, only 1 instance of program allowed
go back to other terminal
type x hit enter
program exits
go back to other terminal once more
./main
No other instance running
x
Last edited by sl4nted; 12-09-2006 at 09:52 PM.
It's exactly what I need.
THANX!!
