Hi all,

I needed help in writing a script for the following question.please help.

Non-recursive shell script that accepts any number of arguments and prints
them in the Reverse order. (For example, if the script is named rargs, then
executing rargs A B C should produce C B A on the standard output).

Thanks

Why can't we use recursion?

#!/bin/sh

case $# in
  0|1) echo "$@";;
  *) arg=$1; shift; exec "$0" "$@" "$arg"
esac

... Actually that's pretty broken, but you get the idea (-:

Without any recursive programming was an interesting twist. But, I figured there had to be "some" reason for this requirement.

Anyway, I assumed by print in reverse order you meant the reverse of entry order. Saying that, perhaps the following script:

#! /bin/bash
#shift_input

tmp_file="shift_value_temp"
tmp_file2="shift_value_temp2"
tmp_file3="shift_value_temp3"
tmp_file4="shift_value_temp4"
rm $tmp_file 2>/dev/null
rm $tmp_file2 2>/dev/null
rm $tmp_file3 2>/dev/null
rm $tmp_file4 2>/dev/null

echo "$*" >$tmp_file
cat $tmp_file | tr " " "\n" >$tmp_file2
tac $tmp_file2 >$tmp_file3
cat $tmp_file3 | tr "\n" " ">$tmp_file4

cat $tmp_file4
echo " "

> shift_input a b c g h i d e f
f e d i h g c b a  
>

Here's one in bash

while [ $# -gt 0 ]
do
  rev="$1 $rev"
  shift
done
echo $rev

while(@ARGV)
{
  unshift @rev, shift @ARGV;
}
print join ' ', @rev, "\n";

For the Perl version, you can just say reverse @ARGV -- no need to code the loop by hand.

#!/bin/sh
exec perl -le '$, = " "; print reverse @ARGV' "$@"

echo "$@" | tr -s " \t" "\n" | tac | tr "\n" " "