How to access command line arguments of the caller inside a function?

Question

I\'m attempting to write a function in bash that will access the scripts command line arguments, but they are replaced with the positional arguments to the function. Is ther

Answer 1

#!/usr/bin/env bash

echo name of script is $0
echo first argument is $1
echo second argument is $2
echo seventeenth argument is $17
echo number of arguments is $#

Edit: please see my comment on question