You wouldn't have had this problem had you not disabled strictures by omitting use strict;. Use use strict;!!!

You did $$hash_ref{$key} in one place and $hash_ref{$key} in the other. The latter accesses the (non-existant) hash %hash_ref, not the hash referenced by $hash_ref.

By the way, $hash_ref->{$key} is a clearer way or writing $$hash_ref{$key}.

And since you already have $hash_ref->{$key} in $val, @{ $hash_ref->{$key} } can be written as @{$val} and @$val.

Cleaned up code: 

#!/usr/bin/perl

use strict;
use warnings;

my $hash_ref = set_up();

foreach my $key (keys %{$hash_ref}) {
   my $val = $hash_ref->{$key};
   if (ref($val)) {
      print "total '$key' = ", scalar @{$val}, "\n";
      foreach my $act (@{$val}) {
         print "\t$act\n";
      }
   } else {
      print "$key - $val\n";
   }
}

sub set_up {
   return {
      one     => "cat",
      two     => "dog",
      actions => [ "feed", "walk", "pet", "groom" ],
   };
}

[download]

In case you don't know, Data::Dumper is your friend when you need to debug a data structure such as the one you are building.

Update: Added code.

In reply to Re: passing hash refs by ikegami
in thread passing hash refs by fredsnertz

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