Is this correct? Daemonizing an app

I am sorry, this may be a stupid question.


I am trying to make a very simple application become a daemon but it seems I can't work it out. I tried to "ps-ef" the application but it returns nothing.

The "daemonizing" code is below.

Code:

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// App.c
#include <stdio.h>

int daemon_init(const char * parAppName){
    int    iCounter;
    int    iPid;
    char  *pTmp;

    // if the fork() call fails, this service won't be started
    if( (iPid = fork()) != 0 ){
        printf("%s Daemonizing failed 1... \n", parAppName);
        exit(0);
    }
   
    // set session id
    setsid();

    //signal( SIGHUP, SIG_IGN);

    // if the fork() call fails, this service won't be started
    if( (iPid = fork()) !=0 ){
        printf("%s Daemonizing failed 2... \n", parAppName);
        exit(0);
    }

    chdir("/");

    umask(0);

    printf("%s Daemonizing Complete... \n", parAppName);

    return 0;
}


int main( int argc, char * argv[] ){
    daemon_init(argv[0]);
    return 0;
}

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when I try to run the program, it results to this :

App Daemonizing Complete...
App Daemonizing failed 2...
App Daemonizing failed 1...

fork() returns -1, AND NO CHILD when it fails.

If it succeeds, then there are TWO processes, one which returns 0 (the child), and one which returns a pid (the parent).

Oh, and daemons usually end up in some kind of infinite loop, waiting to complete their mission.

Code:

---

while ( 1 ) {
  // wait for something to happen
  // do it
}

---

Quote:

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Originally Posted by Salem

fork() returns -1, AND NO CHILD when it fails.

If it succeeds, then there are TWO processes, one which returns 0 (the child), and one which returns a pid (the parent).

Oh, and daemons usually end up in some kind of infinite loop, waiting to complete their mission.

Code:

---

while ( 1 ) {
  // wait for something to happen
  // do it
}

---

---

I got it! Thanks a lot for answering. =)