Neat failure, thanks for posting. Here's a try at an explain. That $var is compiled prior to test(), thus test() can refer to it (you can't move $var after test()), acting as a closure, as has been noted (++). However, as other posters have noted, line 6 does not get executed, so the assignment is never performed. The value of $var is undefined when printed, as a result.

Now, I actually rather believe that storage is allocated on-the-fly, during the call to test(), rather than at compilation-time.

Also, here's some additional test code I wroteup just to check some of my thoughts.

use strict;

&f; #"pre=> ", "post=>2"
my $v = 1;
&f; #pre=>1", "post=>3"
exit;

sub f {
  print "pre=>$v\n";
  $v += 2;
  print "post=>$v\n";
}
[download]

Here's some further code that was illustrative to me.

#this code proves that scratchpads keep lexicals around,
#not mere copies of those lexicals; something like reference
#counting of lexicals is occuring with closures - the details
#don't really matter - the $v lexical persists in both the f1
#and f2 closures, not a mere "copy" of $v.

sub f_maker {
  my $v = 1;
  my $f1 = sub {
    print "f1_pre =>$v\n";
    $v++;
    print "f1_post=>$v\n";
  };
  my $f2 = sub {
    print "f2_pre =>$v\n";
    $v++;
    print "f2_post=>$v\n";
  };
  return ($f1, $f2);
}

my ($f1, $f2) = &f_maker;
&$f1;
&$f2;
&$f1;

__OUTPUT__
f1_pre =>1
f1_post=>2
f2_pre =>2
f2_post=>3
f1_pre =>3
f1_post=>4
[download]