nsluis has asked for the wisdom of the Perl Monks concerning the following question:

Hi! I was playing around with arrays and came across something I didn't expect: @first = ( '1','2','3', '11','22','33' ); @second = @first; $first[0]2 = 4; print "$second[0]2\n"; The result will be 4. I don't want that :) How can I create a copy of the first array without them being 'linked'? TIA, Nanne

Replies are listed 'Best First'.
Re: Creating a copy of an array
by chipmunk (Parson) on Dec 19, 2000 at 03:26 UTC
    How do I print out or copy a recursive data structure?, from perlfaq4, answers this question. However, the docs on this site are a little out of date; this particular FAQ has been updated in newer versions. Use `perldoc -q copy` to look up the answer locally, if you have a more recent version of perl than 5.005_02, or check the docs at http://www.perl.com.

    As to why you had this problem... @second = @first only does a shallow copy. The references in @first are copied to @second, so that @{$second[0]} is the exact same array as @{$first[0]}.

[reply]
Re: Creating a copy of an array
by merlyn (Sage) on Dec 19, 2000 at 04:41 UTC
[reply]
Re:Creating a copy of an array
by I0 (Priest) on Dec 19, 2000 at 04:39 UTC 
    @second = map{[@_]} @first;
#or more generally
use Storable(dclone);
@second = @{dclone(\@first)};

    [download]
[reply]
[d/l]
Re: Creating a copy of an array
by ichimunki (Priest) on Dec 19, 2000 at 04:38 UTC 
    my @first = ( ['1','2','3'],
           ['11','22','33'] );

my @second = map { reref ( @$_ ) } @first;
$first[0][2] = 4;

print "$first[0][2]\n";
print "$second[0][2]\n";

sub reref {
    my @out = ();
    push (@out, $_) for @_; 
    return \@out;
}

    [download]

    This builds a new anonymous list reference for each list in the list. There is probably a better solution, but this works for your stated problem.

    For the record, I have decided to purge the word array from my Perl vocabulary. Perl's lists have only the slightest relationship to arrays in other languages.
[reply]
[d/l]
      How about: my @second = map { [ @$_ ] ] @first; That expands each element then creates a new array ref from it. Without the slowdown of a subroutine call like you used.

      Still there are modules that can do this better.

      Hell you could do this (not recommended, there are modules made for copying). 

      use Data::Dumper qw/Dumper/;
my @first = ( ['1','2','3'],
             ['11','22','33'] );
my @second = @{ eval Dumper(\@first) };

      [download]
[reply]
[d/l]
[select]
        Cool. I think using turning the sub into an expression is much better. So how does any module beat a one liner? :)
[reply]

Back to Seekers of Perl Wisdom