$, = " ";
$d[3] = [qw/just another perl hacker/];
$copy1 = $d[3];
$copy2 = [@{$d[3]}];
print "\$d[3] was", @{$d[3]}, "\n";
$d[3][2] =~ y/p/P/;
print "\$d[3] now", @{$d[3]}, "\n";
print "\$copy1 is", @$copy1, "\n";
print "\$copy2 is", @$copy2, "\n";
__END__
$d[3] was just another perl hacker
$d[3] now just another Perl hacker
$copy1 is just another Perl hacker
$copy2 is just another perl hacker
[download]

well, it works in my case..

insaniac][amano: ~ : perl -we 'use strict;my @d=([0,"BE"],[3,"BUS"],[4
+,"BUS2"]);my @a=map($_ , split(/,/, "BUS2,BE") ) ; my @b=map { my $ma
+pkey=$_; map { $d[$mapkey]->[1] =~ /$a[$_]/ ? $d[$mapkey] : () } 0..$
+#a } 0..$#d; use Data::Dumper;print Dumper(@b);'
$VAR1 = [
          0,
          'BE'
        ];
$VAR2 = [
          4,
          'BUS2'
        ];

insaniac][amano: ~ : perl -we 'use strict;my @d=([0,"BE"],[3,"BUS"],[4
+,"BUS2"]);my @a=map($_ , split(/,/, "BUS2,BE") ) ; my @b=map { my $ma
+pkey=$_; map { $d[$mapkey]->[1] =~ /$a[$_]/ ? [@{$d[$mapkey]}] : () }
+ 0..$#a } 0..$#d; use Data::Dumper;print Dumper(@b);'
$VAR1 = [
          0,
          'BE'
        ];
$VAR2 = [
          4,
          'BUS2'
        ];
[download]

--
to ask a question is a moment of shame
to remain ignorant is a lifelong shame

Add @d to the Dumper() call, and you'll see the difference:

$ perl -we 'use strict;my @d=([0,"BE"],[3,"BUS"],[4,
"BUS2"]);my @a=map($_ , split(/,/, "BUS2,BE") ) ; my @b=map { my $mapk
+ey
=$_; map { $d[$mapkey]->[1] =~ /$a[$_]/ ? $d[$mapkey] : () } 0..$#a } 
+0
..$#d; use Data::Dumper;print Dumper(@d,@b)'
$VAR1 = [
          0,
          'BE'
        ];
$VAR2 = [
          3,
          'BUS'
        ];
$VAR3 = [
          4,
          'BUS2'
        ];
$VAR4 = $VAR1;
$VAR5 = $VAR3;

$ perl -we 'use strict;my @d=([0,"BE"],[3,"BUS"],[4,
"BUS2"]);my @a=map($_ , split(/,/, "BUS2,BE") ) ; my @b=map { my $mapk
+ey
=$_; map { $d[$mapkey]->[1] =~ /$a[$_]/ ? [@{$d[$mapkey]}] : () } 0..$
+#a } 0
..$#d; use Data::Dumper;print Dumper(@d,@b)'
$VAR1 = [
          0,
          'BE'
        ];
$VAR2 = [
          3,
          'BUS'
        ];
$VAR3 = [
          4,
          'BUS2'
        ];
$VAR4 = [
          0,
          'BE'
        ];
$VAR5 = [
          4,
          'BUS2'
        ];
[download]

In the second case, @b is composed of refs to new arrays that have copies of the arrays referenced in @d. People often shoot themselves in the foot by leaving shared references like this in datastructures without meaning to. If, instead of 'BUS2' et. al., you had references, forming an even deeper structure, those would still be shared, though. If you need to prevent that, use something like Storable::dclone($ref) instead of [@{$ref}].

That's not necessarily true. Doing this creates a new arrayref with the same contents, but if down the line he's manipulating the map'd contents but wants to be able to run more results through the same process he may very well want to have copied the original data (since if he alters the map'd contents down the line it'll affect what he's using to build the new list). It is more inefficient than just using the arrayref, but depending on what he's trying to do it might be correct.

again /me crawls away with his head as red as a tomato

Nothing to be ashamed of. You're learning. We all have to do that.

yeah but it kinda bites my ass when i see that i could have found the answer myself :-)
--
to ask a question is a moment of shame
to remain ignorant is a lifelong shame

#!/usr/bin/perl

use strict;
use warnings;  

my @d      = ([0, 1, 2], [3, 4, 5], [6, 7, 8]);
my $mapkey = 1;
    
sub show {
    local($") = ", ";
    print("[", join(", ", map({"[@$_]"} @d)), "]\n");
}

my $one    = [@{$d[$mapkey]}];
my $two    = $d[$mapkey];

                   show;  # Show array. Unmodified.
$one->[1] = "foo"; show;  # Array is still unmodified.
$two->[1] = "foo"; show;  # This modifies the array.
 
__END__
[[0, 1, 2], [3, 4, 5], [6, 7, 8]]
[[0, 1, 2], [3, 4, 5], [6, 7, 8]]
[[0, 1, 2], [3, foo, 5], [6, 7, 8]]
[download]