in reply to Copy of multidimensional hash

References are reference :) a multidimensional hash stores references, a hash of arrays stores array references as values(hash value) associated with hash-keys, and the same ones refers to the same value
$ perl -le " %F =(1, [41] ); $F{2}= $F{1}; print for %F; " 1 ARRAY(0x3f9074) 2 ARRAY(0x3f9074)

value of 1 is same as 2, same reference ARRAY(0x3f9074), same array, so  $F{2}[0]++ makes  $F{1}[0] 42, because $F{1} and $F{2} refer to the same array

See Tutorals: References, and perlref

Replies are listed 'Best First'.
Re^2: Copy of multidimensional hash
by tommaso.fornaciari (Initiate) on Jan 03, 2013 at 09:19 UTC

    Dear All Monks,
    Thanks a lot for your quick and clear answers!
    In fact, storing elsewhere the same data, it works:

    #!/usr/bin/env perl use Modern::Perl 2011; use autodie; # I create a multidimensional hash my %one = ('a' => [1, 2], 'b' => [3, 4], 'c' => [5, 6]); # I create a copy: # wrong way: #my %two = %one; # right way: my %two; for(keys %one) {$two{$_}[0] = $one{$_}[0]; $two{$_}[1] = $one{$_}[1]} # I print them: say "Before:\n\%one:"; for(sort keys %one){say "$_\t$one{$_}[0]\t$one{$_}[1]"} say '%two:'; for(sort keys %two){say "$_\t$two{$_}[0]\t$two{$_}[1]"} # Then I modify the copy: for(sort keys %two){$two{$_}[1] = 'two'} # And I print again: say "After:\n\%one:"; for(sort keys %one){say "$_\t$one{$_}[0]\t$one{$_}[1]\tOk!!!"} say '%two:'; for(sort keys %two){say "$_\t$two{$_}[0]\t$two{$_}[1]"} # I receive this output: #Before: #%one: #a 1 2 #b 3 4 #c 5 6 #%two: #a 1 2 #b 3 4 #c 5 6 #After: #%one: #a 1 2 Ok!!! #b 3 4 Ok!!! #c 5 6 Ok!!! #%two: #a 1 two #b 3 two #c 5 two

    Is this the best way to create a copy of a multidimensional hash?
    Thanks again for your helpfulness!
    Tommaso

      Is this the best way to create a copy of a multidimensional hash?

      I don't know if it's the best. At any rate, I would reformat the for loop as follows:

      for (keys %one) { $two{$_}[0] = $one{$_}[0]; $two{$_}[1] = $one{$_}[1]; }

      I find this significantly more readable than the way you formatted it. Furthermore, I would generalize copying the arrray, and tell Perl to take out all the elements instead of just those elements with index 0 and 1.

      for my $k (keys %one) { $two{$k}[$_] = $one{$k}[$_] for 0..scalar( @{$one{$k}} ); }

      And then I would generalize it even further: take the whole array reference, dereference it in a single pass, and create and store a new reference to it. Sounds complex? Nah, not so much:

      for my $k (keys %one) { $two{$k} = [ @{$one{$k}} ]; }

      This was already suggested by johngg.</p

      Step by step of [ @{$one{$k}} ]:

      1. $one{$k}: this takes the element identified by key $k from hash %one. That element, in this case, is an array reference.
      2. @{ ... } dereferences the array reference inside the curlies. So after @{ $one{$k} } we're working with a normal, regular, every day array (albeit an anonymous one).
      3. [ ... ] stores whatever is between the square braces as an anonymous array and returns a reference to it.

      But then I would want to generalize it even further and use Storable's dclone function, as suggested by davido.

      use Storable qw(dclone); ... my %two = %{ dclone(\%one) } # dclone() takes a ref and returns a r +ef # So we'll have to put in \%one (not % +one) # and then we have to dereference ( %{ +...} ) # what comes out.

        Thanks muba, thanks johnGG,
        great dissertation about the the topic :))

      You could use Clone and copy the hash using my %two = %{ Clone::clone(\%one) };
      Or use hashrefs all the way:
      my $hr_one = { 'a' => [1, 2], 'b' => [3, 4], 'c' => [5, 6]}; my $hr_two = Clone::clone($hr_one);