How do i print arguments in reverse order in shell script? [duplicate]

Answer 1

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.

Answer 2

You need eval with echo i.e. you need to evaluate the expansion, not output it:

eval echo $a$(($#-i))

Note that, using eval in general is discouraged as this could result in security implications if the input string is not sanitized. Check John1024's answer to see how this can be done without eval.