Convert command line arguments into an array in Bash

Question

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 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

Here is another usage :

#!/bin/bash
array=( "$@" )
arraylength=${#array[@]}
for (( i=0; i<${arraylength}; i++ ));
do
   echo "${array[$i]}"
done