Both calculations are done based on the structure member which holds
the highest used index in the array. Evaluating the array in scalar
context is actually less expensive for two reason:

        1. In Perl versions prio to 5.12, no intermediate 'magic SV' is
           created because the result of this evaluation is not an
           lvalue.

        2. Calculating the number of elements is done by adding 1 to
           the 'highest used index' value. But calcluating $#array
           requires adding the current 'first index' value ($[, =>
           perval(3pm)).

NB: On the computer where I tested this, the difference was about
0.000002s (2E-6).

-------------------
use Benchmark;

my @a = 1 .. 200;

timethese(-5,
          {
           len => sub {
               my $l;

               $l = @a;
               return $l;
           },

           last => sub {
               my $l;