Sorry. I made an unreasonable assumption.

Edit: I took a look at your source, and I'm unsure about the reasons you're doing things the way you are. First, shell scripts should have their first line set to ``#!/bin/sh'' (or whatever shell you want to use), and should be set as executable. Second, it seems like a waste of resources to start bash via execve() in order for it to just execute other commands, unless you want to be able to execute commands like ``cd /drive0/var/emptris && ./emptris'', which would, in all honesty, probably solve your problem here.

My guess is that you're trying to execute startemptris by execve()ing ``/bin/sh -c startemptris''. First, the `-c' tells bash that the arg following is a shell command, instead of a script. Second, startemptris is probably not in bash's search path, so you'd need to specify a path to startemptris, even if it's just ``./startemptris''. Third, I'm guessing that startemptris isn't executable, so it'll still fail. My guess is that you should try to start the programs without necessarily executing bash to start them. Just run the programs directly and make sure that they're properly configured. That way, your program could run both shell scripts and binary executables, plus you avoid that useless overhead.

Also, you don't need to `&' empegVNC; it backgrounds itself, or at least the newer versions do.