Question

Here is my code

#include <sys/types.h>
#include <sys/stat.h>
#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <errno.h>
#include <unistd.h>
#include <syslog.h>
#include <string.h>
#include <iostream>

int main(int argc, char *argv[]) {
    pid_t pid, sid;
    int sec = 10;

    pid = fork();
    if (pid < 0) {
        perror("fork");
        exit(EXIT_FAILURE);
    }
    if (pid > 0) {
        std::cout << "Running with PID: " << pid << std::endl;      
        exit(EXIT_SUCCESS);
    }
    umask(0);        

    sid = setsid();
    if (sid < 0)
        exit(EXIT_FAILURE);

    if ((chdir("/")) < 0)
        exit(EXIT_FAILURE);    /* Log the failure */

    close(STDIN_FILENO);
    close(STDOUT_FILENO);
    close(STDERR_FILENO);

    while (1) {
        execl("/bin/notify-send", "notify-send", "-t", "3000", "Title", "body", NULL);
        sleep(sec); 
    }
    exit(EXIT_SUCCESS);
}

I want it to give a notification in every 10 sec. Daemon is running OK, but not giving any notification.

Answer 1

execl doesn’t return – it replaces the running program with a new one. So running it in a loop with sleep is meaningless – it will run only once.

I think you should use system instead. It executes a command and returns.

The alternative is to fork each time in the loop, and have the child do execl while the parent would continue the loop.