The list of

(1 => 2, 3 => $x{1} )
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(1, 2, 3, undef)
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In fact, it's the prior value of $x{1} that you see, as evidenced by:

%x{1} = "old x sub 1";
%x = 
  ( 1 => 2, 
    3 => $x{1} 
  );
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The question for me is, why did you find this surprising or baffling? It's extremely consistent with what Perl does at all times: evaluate the right side of the assignment before altering any structure on the left. It's what permits

@a = (2, @a)
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@a = reverse sort @a;
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-- Randal L. Schwartz, Perl hacker

    $x = ($x + 18) / 7;
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Are you kidding? Mathematical expansion like that would kick ass.

japhy -- Perl and Regex Hacker

I'm not kidding. Some languages have something a little like that. But you quickly get into big problems. Consider:

    my $x = not $x;
    my $y = $y-1;
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or

    my @a = (@a, 3);
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for example.

CODE:

use Data::Dumper;

%x =
  ( 3 => 4,
    1 => 5
  );

print Dumper(\%x), $/;

%x =
  (  1 => 2,
     3 => $x{1}
  );


print Dumper(\%x), $/;
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$VAR1 = {
          1 => 5,
          3 => 4
        };

$VAR1 = {
          1 => 2,
          3 => 5
        };
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Something I don't know about though is if there is a notation for within an anonymous hash to have it be able to reference itself. That way you *could* use this behavior to set other values based on itself while it is building. I also remember reading a while back that hashes are constructed backwards (somebody would have to verify this because I'm not 100% sure) so it is possible that even if you could do this it would probably be a little confusing.

I'm not sure what you are getting at about the %y values being immediately bound after each statement is run because if this behavior didn't exist it would be impossible to code in perl because everything would be contingent on side effects which is a rather scary thought. After each semi-colon perl really ought to have moved all the bits around in the way I told it to.

Cool find though... I certainly never would have bothered to try and access a hash's data while constructing the hash itself. On the other hand it probably isn't the best coding practice either, but what the hell!

As I see it, this has little to do with hashes at all. Simply put it is a bit like trying to declare two variables thus:

my ($a,$b)=(25,$a/5);
print "$a and $b\n";

Hopefully that looks immediately wrong. For the reasons everyone has indicated above.

--

%hash = (
  a => 1,
  b => 2,
  c => $_{a},
  d => $_{b},
);
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I'll jump right to the elegant solution to japhy's question: RE: multiple keys - one value

        - tye (but my friends call me "Stupendous Search")

In your later example, you are refering to an already existing $y{1}, so you can access it's contents.

=Blue
...you might be eaten by a grue...