How do I write a watchdog daemon in bash?

Based on clarifications in comments, what you actually want is a daemon process that keeps a child running, relaunching it whenever it exits. You want a way to type "./myscript.sh" in an ssh session and have the daemon started.

#!/usr/bin/env bashPIDFILE=~/.mydaemon.pidif [ x"$1" = x-daemon ]; then  if test -f "$PIDFILE"; then exit; fi  echo $$ > "$PIDFILE"  trap "rm '$PIDFILE'" EXIT SIGTERM  while true; do    #launch your app here    /usr/bin/server-or-whatever &    wait # needed for trap to work  doneelif [ x"$1" = x-stop ]; then  kill `cat "$PIDFILE"`else  nohup "$0" -daemonfi

Run the script: it will launch the daemon process for you with nohup. The daemon process is a loop that watches for the child to exit, and relaunches it when it does.

To control the daemon, there's a -stop argument the script can take that will kill the daemon. Look at examples in your system's init scripts for more complete examples with better error checking.

The pid of the most recently "backgrounded" process is stored in $!

$ cat &[1] 7057$ echo $!7057

I am unaware of a fork command in bash. Are you sure bash is the right tool for this job?