subroutine reference questions

convenientstore has asked for the wisdom of the Perl Monks concerning the following question:

Hello, while reading Intermediate perl, I got stuck on page 97. which has this code

#!/usr/bin/perl -w

use strict;
use Data::Dumper;

sub data_for_path {
  my $path = shift;
  if (-f $path or -l $path) {     #files or symbolic links
   return undef;
  }
  if (-d $path) {
    my %directory;
    opendir PATH, $path or die "cannot opendir $path: $!";
    my @names = readdir PATH;
    closedir PATH;
    for my $name (@names) {
     next if $name eq '.' or $name eq '..';
     $directory{$name} = data_for_path("$path/$name");
    }
    return \%directory;
  }
  warn "$path is neither a file nor a directory\n";
  return undef;
}

print Dumper(data_for_path('.'));
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which shows me the output of

$VAR1 = {
          'perl.1' => undef,
          'bin' => {
                     'perl.1' => undef,
                     'chaos.1' => undef
                   },
          'perl.sub_sub' => undef,
          'perl.sub_multi' => undef,
          'chaos.1' => undef,
          'perl.filehandle' => undef,
          'man' => {},
          'lib' => {
                     'yahoo3' => {
                                   'yahoo3' => {
                                                 'yahoo1111' => undef
                                               },
                                   'yahoo5' => undef
                                 },
                     'yahoo2' => undef,
                     'yahoo' => undef
                   },
          'perl.log' => undef,
          'perl.recursive_data' => undef,
          'perl.callbacks' => undef
        };
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But I was trying to do it without datadumper to see if I can print out the structure..(possibly with references) tried couple things but does not work.. any suggestion? Problem I see is that I am not so sure what is being return from the sub routines.. and even if I know what it looks like, if sub directory has multiple subdirectories.. woudln't it make it almost impossible to build a right structure of reference?

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Re: subroutine reference questions by brian_d_foy (Abbot) on Sep 01, 2007 at 20:17 UTC
You have to look on the next page. :) In "Displaying Recursively Defined Data" on page 98, we show you how to do it. The `dump_data_for_path` routine, which is also recursive. displays the data structure. The code itself is on page 99: `sub dump_data_for_path { my $path = shift; my $data = shift; if (not defined $data) { # plain file print "$path\n"; return; } my %directory = %$data; for (sort keys %directory) { dump_data_for_path("$path/$_", $directory{$_}); } }` [download] -- brian d foy <brian@stonehenge.com> Subscribe to The Perl Review	[reply] [d/l] [select]
Re^2: subroutine reference questions by convenientstore (Pilgrim) on Sep 01, 2007 at 23:32 UTC
:-) thanks~	[reply]
Re: subroutine reference questions by FunkyMonk (Bishop) on Sep 01, 2007 at 18:53 UTC
Have a look at ref. It'll tell you what kind of reference the argument is, or return an empty string if it's a scalar. For example: `my $hashref = {}; my $arrayref = []; my $coderef = sub {}; my $scalar = ''; print "[", ref, "]\n" for $hashref, $arrayref, $coderef, $scalar;` [download] Outputs `[HASH] [ARRAY] [CODE] []` [download]	[reply] [d/l] [select]