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

Reversing a simple string, by spaces

Simply:

#!/bin/sh
o=
for i;do
    o="$i $o"
    done
echo "$o"

will work as

./rev.sh 1 2 3 4
4 3 2 1

Or

./rev.sh world! Hello
Hello world!

If you need to output one line by argument

Just replace echo by printf "%s\n":

#!/bin/sh
o=
for i;do
    o="$i $o"
    done

printf "%s\n" $o

Reversing an array of strings

If your argument could contain spaces, you could use bash arrays:

#!/bin/bash

declare -a o=()

for i;do
    o=("$i" "${o[@]}")
    done

printf "%s\n" "${o[@]}"

Sample:

./rev.sh "Hello world" print will this
this
will
print
Hello world