perl - ディレクトリ構造をコピー元からコピー先にコピーする perl のスクリプト

Question

#!/usr/bin/perl -w
use    File::Copy;
use    strict;

my   $i=  "0";
my   $j=  "1";
my   $source_directory = $ARGV[$i]; 
my    $target_directory = $ARGV[$j];
#print $source_directory,"\n";
#print $target_directory,"\n";

my@list=process_files ($source_directory);
print "remaninign files\n";
print @list;
# Accepts one argument: the full path to a directory.
# Returns: A list of files that reside in that path.
sub process_files {
    my $path = shift;

    opendir (DIR, $path)
     or die "Unable to open $path: $!";

    # We are just chaining the grep and map from
    # the previous example.
    # You'll see this often, so pay attention ;)
    # This is the same as:
    # LIST = map(EXP, grep(EXP, readdir()))
    my @files = 
        # Third: Prepend the full path
        map { $path . '/' . $_}
        # Second: take out '.' and '..'
        grep { !/^\.{1,2}$/ }
        # First: get all files
        readdir (DIR);

    closedir (DIR);

    for (@files) {
        if (-d $_) {
            # Add all of the new files from this directory
            # (and its subdirectories, and so on... if any)
            push @files, process_files ($_);

        } else { #print @files,"\n";
          # for(@files)
        while(@files) 
        {  
        my $input= pop @files;
        print $input,"\n";
        copy($input,$target_directory);

             }
  }
      # NOTE: we're returning the list of files
   return @files;
  }
}

これは基本的にソースから宛先にファイルをコピーしますが、ディレクトリをコピーする方法についてもガイダンスが必要です。ここで注意すべき主なことは、copy、move、および path 以外の CPAN モジュールが許可されていないことです。