This is straightforward to predict when you separate what's happening on the right side from what's happening on the left.

The list of

(1 => 2, 3 => $x{1} )
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Is identical to

(1, 2, 3, undef)
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as long as %x didn't have a previous value.

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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or

@a = reverse sort @a;
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to work.

-- Randal L. Schwartz, Perl hacker