it's equivalent to

 
push @{ 
    if( $count{$element} > 1 )
    {
        \@intersection;
    }
    else
    {
        \@difference;
    }
}, $element;
[download]

~Particle *accelerates*

# initialize all sets to be empty
@union = @intersection = @difference = ();
%count = ();

# count the number of times each unique element appears
# in both arrays
foreach $element (@array1, @array2) { $count{$element}++ }

# for each unique element, determine whether it belongs
# to the union, intersection, or difference of the original arrays
foreach $element (keys %count) {

    # since the union includes all elements present in either
    # array, every unique element should belong to the union
    push @union, $element;

    # if this element occurred appeared more than once
    # (i.e., in both arrays), then it belongs to the intersection.
    # if it occurred only once (i.e., in only one array), then it
    # belongs to the difference.  perlfaq4 explains that this is
    # -symmetric difference-, like an XOR operation.
    push @{ $count{$element} > 1 ? \@intersection : \@difference }, $e
+lement;
}
[download]

It seems to me, though, that this code only works if the elements of each array are unique. If the same element appears twice in one array, it will become part of the intersection set, although it shouldn't. (Update: If I had read perlfaq4 carefully myself, I would have seen that this assumption is documented. oops :-)

@{ ... } is a dereferencing operation. It's useful to note that the { ... } is just a normal code block and can contain any code you want, as long as it returns an array reference.

For instance, you could call a subroutine which returns an array reference:

 foreach my $case ( $justice_wheels->spin() ) {
     my $verdict = Conviction::Felony->new( $case->defendant() );
     push @{ $case->defendant()->get_felony_convictions() }, $verdict;
     push @{ $case->judge()->get_felony_convictions() }, $verdict;
 }
[download]

Yep, definitely feeling silly today...

Chris
M-x auto-bs-mode

Ahhh... pretty cool, I'll be using ? definitely more in the future. Thanks!