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.

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

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 aa12 2 bb12 3 cc12 4 dd12 5 5512 6 ff12 7 gg12 8 hh12 9 ii12 10 jjj12 11 kkk12 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 defaultarg2fi

which translates to this even simpler expression:

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