Yeah it's best to just use POSIX shell arithmetic, and not be clever, and not the other 2 ways bash has to do it! THat is: i=$((i + 1)) The (( )) constru

Yeah it's best to just use POSIX shell arithmetic, and not be clever, and not the other 2 ways bash has to do it! THat is: i=$((i + 1)) The (( )) construct, without the $, is different, and it's also not POSIX. It has the gotchas around the exit code, which a plain assignment doesn't have.