in reply to Elegantly dereferencing multiple references

FWIW it's possible in one statement but hardly "elegant" 
use strict;
use warnings;
use Data::Dump;

my %x = ( a  => "red");
my %y = ( b  => "green");
my %z = ( c  => "black");

sub assign {
   my (%x, %y, %z) = ( %{$_[0]}, %{$_[1]}, %{$_[2]} );
   dd  \(%x, %y, %z);
}

[download]

I'll post later a more elegant but significantly slower solution

UPDATE:
I was wrong, never mind. Thanks to Eily++

Cheers Rolf
(addicted to the Perl Programming Language and ☆☆☆☆ :)
Je suis Charlie!

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Re^2: Elegantly dereferencing multiple references
by Eily (Monsignor) on May 13, 2016 at 12:31 UTC

    I was wondering what trick you used to prevent the flattened list to go into the first hash only. Looks like there's none: 

    $VAR1 = {
          'c' => 'black',
          'a' => 'red',
          'b' => 'green'
        };
$VAR2 = {};
$VAR3 = {};

    [download]

[reply]
[d/l]
      oops you are right, I had various outputs from Data::Dump and only saw what I expected to see... :-/

      Cheers Rolf
      (addicted to the Perl Programming Language and ☆☆☆☆ :)
      Je suis Charlie!

[reply]
Re^2: Elegantly dereferencing multiple references
by LanX (Saint) on May 13, 2016 at 13:44 UTC
    as compensation an even less elegant (read uglier) attempt for a one liner:

    sub assign {
   %$_ = %{shift @_} for \ (my (%a, %b, %c));
   dd ["assign", \(%a, %b, %c) ];
}
assign( \(%x, %y, %z));

    [download]

    Cheers Rolf
    (addicted to the Perl Programming Language and ☆☆☆☆ :)
    Je suis Charlie!

[reply]
[d/l]

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