What is a reason for using shift $((OPTIND-1)) after getopts?

Question

I realise that shift moves the array of cli args n space to the left, and the default of n is 1. This means I can assign the values the array to existing varibles using $1 shift inside a while loop. What I don't quite understand is why it is used in this context below. The input args have been assigned to values already and deleting shift $((OPTIND-1)) doesnt change this fact. Source: http://linux.die.net/man/3/optind

while getopts ":h:a:fc" opt; do
    case $opt in
        h)
            print_help
            exit 0
            ;;
        a)
            aaaa=${OPTARG}
            ;;
        f)
            force=1
            ;;
        c)
            CLEAN=1
            ;;
        \?)
            echoerr "Invalid option -$OPTARG"
            print_help
            exit 1
            ;;
    esac
done

shift $((OPTIND-1))

Answer 1

The shift removes the parameters processed by the getopts loop from the parameter list, so that the rest of the script can process the remainder of the command line (if any) with $1... in the usual manner, without concern for the number of options processed by getopts.

Consider a hypothetical script with the usage

frobble [-f] [-c] [-a hostname] filename ...

The getopts loop above takes care of parsing the -f,-c and -a, if they are present, but doesn't remove them from the argument list. Which means that to get at your filename argument, you need to find out how many options were processed, and continue processing from there. Conveniently, getopts tells you the index of the first unprocessed argument: the variable OPTIND.

And instead of messing with offsets and stuff, you can just discard the processed options, renumbering the rest of the arguments so your filename is always $1.

That is what shift $((OPTIND-1)) does.