Convert command line arguments into an array in Bash

Actually your command line arguments are practically like an array already. At least, you can treat the $@ variable much like an array. That said, you can convert it into an actual array like this:

myArray=( "$@" )

If you just want to type some arguments and feed them into the $@ value, use set:

$ set -- apple banana "kiwi fruit"
$ echo "$#"
3
$ echo "$@"
apple banana kiwi fruit

Understanding how to use the argument structure is particularly useful in POSIX sh, which has nothing else like an array.

Actually the list of parameters could be accessed with $1 $2 ... etc.
Which is exactly equivalent to:

${!i}

So, the list of parameters could be changed with set,
and ${!i} is the correct way to access them:

$ set -- aa bb cc dd 55 ff gg hh ii jjj kkk lll
$ for ((i=0;i<=$#;i++)); do echo "$#" "$i" "${!i}"; done

12 1 aa
12 2 bb
12 3 cc
12 4 dd
12 5 55
12 6 ff
12 7 gg
12 8 hh
12 9 ii
12 10 jjj
12 11 kkk
12 12 lll

For your specific case, this could be used (without the need for arrays), to set the list of arguments when none was given:

if [ "$#" -eq 0 ]; then
    set -- defaultarg1 defaultarg2
fi

which translates to this even simpler expression:

[ "$#" == "0" ] && set -- defaultarg1 defaultarg2

Maybe this can help:

myArray=("$@") 

also you can iterate over arguments by omitting 'in':

for arg; do
   echo "$arg"
done

will be equivalent

for arg in "${myArray[@]}"; do
   echo "$arg"
done