Re: Why Perl boolean expression sometimes treated as lvalue?

Hmmm... interesting disparity. The following may be of interest...

use v5.10;

my ($a, $b) = (0, 0);

sub lv :lvalue { $_[0] }

lv($a || $b) = 2;
lv($a && $b) = 1;

say $a;
say $b;
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package Cow { use Moo; has name => (is => 'lazy', default => sub { 'Mooington' }) } say Cow->new->name