No and yes. No, they are not shared in any way which is visible to the programmer; the processes can modify their own copies of the variables independently and they will change without any noticable effect on the other process(es) which are fork() parents, siblings or descendents. But yes, the OS actually does share the … Read more
Since you mentioned a daemon, I can conclude that you are running on a Unix-like operating system. This matters, because how to do this depends on the kind operating system. This answer applies only to Unix, including Linux, and Mac OS X. Define a function that will set the gid and uid of the running … Read more
For sending the output to another file (I’m leaving out error checking to focus on the important details): if (fork() == 0) { // child int fd = open(file, O_RDWR | O_CREAT, S_IRUSR | S_IWUSR); dup2(fd, 1); // make stdout go to file dup2(fd, 2); // make stderr go to file – you may choose … Read more
More specifically, from the Redis FAQ Redis background saving schema relies on the copy-on-write semantic of fork in modern operating systems: Redis forks (creates a child process) that is an exact copy of the parent. The child process dumps the DB on disk and finally exits. In theory the child should use as much memory … Read more
This answer is probably a little late but: http://akuma.kohsuke.org/ seems to be exactly what your looking for
No. If the parent is killed, children become children of the init process (that has the process id 1 and is launched as the first user process by the kernel). The init process checks periodically for new children, and waits for them (thus freeing resources that are allocated by their return value). The question was … Read more
This isn’t quite what you thought originally. The output buffer is not shared – when you execute the fork, both processes get a copy of the same buffer. So, after you fork, both processes eventually flush the buffer and print the contents to screen separately. This only happens because cout is buffered IO. If you … Read more
In bash scripts (non-interactive) by default JOB CONTROL is disabled so you can’t do the the commands: job, fg, and bg. Here is what works well for me: #!/bin/sh set -m # Enable Job Control for i in `seq 30`; do # start 30 jobs in parallel sleep 3 & done # Wait for all … Read more
Today this is pretty easy to do with Bundler. You make a local copy of the gem and then instead of doing gem “whatever” in your Gemfile, you do: gem “whatever”, :path => “/home/pupeno/whatever” After running bundle install, the gem is picked from that directory. Even if you modify something in there, all you need … Read more
pkill -P $$ will fit (just kills its own descendants) And here is the help of -P -P, –parent ppid,… Only match processes whose parent process ID is listed. and $$ is the process id of the script itself