How would you write a shell script that prints all the additional arguments passed to it in reverse order?

Answer 1

for (( i = ${#}; i > 0; i-- )); do
        echo ${!i}
done

The arguments are available as $, where n is the position of the argument. For example, $0 would give the name of the script, $1 would give the first additional argument, $2 the second, and so on. The total number of additional arguments is found in $#.

A loop that starts at $# and ends at 1 can be used to print each additional argument in reverse order.