Let's define your arguments:
$ set -- one two three
Now, let's print them out in reverse order:
$ for ((i=$#;i>=1;i--)); do echo "${!i}"; done
three
two
one
How it works
for ((i=$#;i>=1;i--)) starts a loop in which i counts down from $# to 1. For each value of i, we print the corresponding positional parameter by ${!i}. The construct ${!i} uses indirection: instead of returning the value of i, ${!i} returns the value of the variable whose name is $i.
As a script
In a multi-line script form, we can use:
$ cat reverse
#!/bin/bash
for ((i=$#;i>=1;i--))
do
echo "${!i}"
done
As an example:
$ bash reverse One Two Three
Three
Two
One
Alternative: using tac
Another way to print things in reverse order is to use the utility tac. Consider this script:
$ cat reverse2
#!/bin/bash
printf "%s\n" "$@" | tac
Here is an example:
$ bash reverse2 Uno Dos Tres
Tres
Dos
Uno
printf "%s\n" "$@" prints out the positional parameters one per line. tac prints those lines in reverse order.
Limitation: The tac method only works correctly if the arguments do not themselves contain newlines.
