Hello all, I am wondering what the best way to write this bash script in perl would be. All the script is doing is getting the current directory, getting a different directory, and then finally printing out all the directories present in that different directory. 
#!/bin/bash
# $1 results directory
# to run, e.g., 
# step1 /your/path/to/results 

# need one argument ! 
if [ $# == 1 ]; then 

        # scripts directory
        d=$( dirname $( readlink -m $0 ) )
        #results directory
        res=$( readlink -m $1 )
        # loop through the cases in data
        for sample in $( ls ${res}/ );do
        echo $sample;
        done;
    else
        echo "error - need to supply an argument"
fi

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What would the best way to write this in perl be ? Is there anyway to avoid using opendir, readdir, and closedir? and to avoid having to use CWD to get the absolute path ( as readlink -m . does in bash, or a way to write the for loop without having to first read into an array?

. The following perl script does the same thing, but I want to streamline it, even if that means really obfuscating it, any suggestions? 
#!/usr/bin/perl -w
use strict;
use Cwd 'abs_path';
use Cwd;
use File::Basename;

if (defined ($ARGV[0])){
#scripts directory
    my $d= getcwd();;
#results directory
    my $res= abs_path($ARGV[0]);
  
 # loop through the cases in data
         opendir( DIR, $res ) || die "Error in opening dir $res\n";
             my @files = readdir(DIR);
         close(DIR);
     foreach my $sample (@files){
            print $sample . "\n";
        }
  }
else {print "Need to supply argument\n"};

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In reply to Most effective bash to perl for directory listing by ZWcarp

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