Updates from one view will be seen from the other.

They will? I tried the following: 

#!/usr/bin/perl -w
use strict;

my %hash;
$hash{foo}{bar}{x} = 'zoom!';
$hash{bar}{foo}{x} = $hash{foo}{bar}{x};

print "$hash{foo}{bar}{x}\n";
print "$hash{bar}{foo}{x}\n";

$hash{foo}{bar}{x} = '!mooz';

print "$hash{foo}{bar}{x}\n";
print "$hash{bar}{foo}{x}\n";

[download]
which outputs: 
zoom!
zoom!
!mooz
zoom!

[download]
Now, if maybe I'm misunderstanding what your code does, but while it gives two separate views of the data, it does not create a duplicate set of references to a single set of data... if my calculations are correct it creates a second copy of the data with new references to the second copy. Changes to one copy will not affect the other.
update: my fault for not paying attention to details! Thanks for clarifying zaxo. Of course, this is still less than optimal, since it would seem to make doing key reordering impossible for that last key (not that doing this was in the original requirements, but it's the kind of solution I'd be hoping for). Also, we do still have a potential key overwrite going on (think $hash{foo}{foo}{x}).

In reply to (ichi) Re x 2: creating multiple views of a hash by reordering its key and values by ichimunki
in thread creating multiple views of a hash by reordering its key and values by princepawn

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