Print bash arguments in reverse order

Question

I have to write a script, which will take all arguments and print them in reverse.

I\'ve made a solution, but find it very bad. Do you have a smarter idea?

Answer 1

Could do this

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

This uses the below
c style for loop
Parameter indirect expansion (${!i}towards the bottom of the page)

And $# which is the number of arguments to the script