shell script

shambhavi · 09-21-2012, 07:14 AM

$ecursive shell script that accepts any number of arguments and prints them in reverse order.
code:

Code:

 len=$#
  while [ $len -ne 0 ]
do
eval echo '$' $len
len=`expr $len - 1`
done

But when i used sh filename.sh hello hi, It is returning

Code:

 $2
 $1

as the output.
How to correct it?

acid_kewpie · 09-21-2012, 07:30 AM

can you please format this correctly using [ code ] tags? I've no idea what is bash and what is your improvised formatting.

rosehosting.com · 09-21-2012, 07:32 AM

This code should do the work:

Code:

#!/bin/bash

#  array from input
_ARRAY=( $* )

#iterate over the array in reverse order.
for ((i=${#_ARRAY[@]}-1; i>=0; i--)); do
  echo "${_ARRAY[$i]}"
done

Code:

./filename.sh 3 5 74 2
2
74
5
3

acid_kewpie · 09-21-2012, 07:34 AM

rosehosting.com, please don't give verbatim answers to homework...

Per the LQ Rules, please do not post homework assignments verbatim. We're happy to assist if you have specific questions or have hit a stumbling point, however. Let us know what you've already tried and what references you have used (including class notes, books, and Google searches) and we'll do our best to help. Also, keep in mind that your instructor might also be an LQ member.

catkin · 09-21-2012, 07:35 AM

Quote:

Originally Posted by shambhavi

How to correct it?

Remove the space from '$' $len

EDIT: that makes your solution work but does not meet the requirement to produce a recursive solution.

David the H. · 09-23-2012, 07:41 AM

I'll just add my two three cents:

1) Never use eval. Not unless you know exactly what you are doing. And if you know what you are doing, you will almost never need it.

2) There is never any good reason to use expr in any modern shell. They all have perfectly good integer arithmetic built in.

3) Indirect/variable variables are also unnecessary in most modern shells, and sloppy and non-transparent in any case. There are better techniques, such as the array solution that rosehosting posted.

konsolebox · 09-23-2012, 08:22 AM

Well it's actually not necessary to use arrays there.

Code:

#!/bin/bash
for (( I = $#; I; --I )); do
    echo "${!I}"
done