So while looking on getopt i fount out that there needs to be an extra case for -- even though i dont want parameters in the format like -flag -flag -- argument1 argument2 etc.Any specific reason for why getopt need a special case for --. Just curious.
Sample Script:-

1 set -- $(getopt -q ab:c "$@")
2 while [[ -n "$1" ]]
3 do
4 case "$1" in
5 -a) echo "Found the -a option" ;;
6 -b) param="$2"
7 echo "Found the -b option, with parameter value $param"
8 shift 2;;
9 -c) echo "Found the -c option" ;;
10 --)shift
11 break;;
12 *) echo "$1 is not an option";;
13 esac
14 shift
15 done

getopt does not need --. But probably I misunderstood your question.
-- is a special flag and is passed to set, not getopt. The meaning of -- is: do not take all the arguments (on the right side of --) as options even if they started with -.
see man bash and look for --

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Sorry if you got confused but i was actually talking about the -- that is present in switch statement. The --) case. But now i understood why we need it there thanks for your help.

getopt itself will add a --. You can check the command:
and you will see the result will contain an additional -- (again, left side the options, right side the additional arguments).
in your case therefore you need to handle that -- too.

Insert set -xv at the beginning of your script and you will see what's really happening.

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Using getopt for a while but was remembering that case. Most of the people advised getopts without explaining the why now i know the why thanks for your help again.