From the perlfunc documentation for each:

There is a single iterator for each hash, shared by all each, keys, and values function calls in the program

Hi,

well - I know of this restriction, and it may do bite me here sometime later. But I do not see how the each iterator, bound to %$rul should clash with the iteration of for(@out).

Bye
 PetaMem     All Perl:   MT, NLP, NLU

Fletch and splinky are right. Try this toy example and you will see it clearly:

use strict;
use warnings;

my $count = 0;
my $max_output = 7;

sub rec {
  my $hash  = shift;
  my $level = shift;
  return if $level > 1;

  while ( my ( $k, $v ) = each %$hash ) {
    exit 0 if ++$count > $max_output;   # backstop
    print "$level: $k\n";
    rec( $hash, $level + 1 );
  }
}

rec( { foo => 1, bar => 2, baz => 3 }, 0 );

__END__
0: bar
1: baz
1: foo
0: bar
1: baz
1: foo
0: bar
[download]

Notice how, when the recursion level is 1, the iterator doesn't visit 'bar' (because it was already visited at level 0). Instead, it visits the remaining keys (thereby exhausting each), and returns (since in this case, all further recursion stops at level 2). Once level 1 returns control to level 0, each starts again from the top.

the lowliest monk