Clear questions and runnable code
get the best and fastest answer
> am I correct in thinking that this is using the or operator as you would like += or .=
almost, you are describing ||= which assigns if the left side is false. you can find this combination in many languages.
But current Perl versions also have // for defined-or, which only act on undef and not other false values.
so $type //= "SCALAR" means $type = defined $type ? $type : "SCALAR";
> performs the operation on $_. Currently I'm trying to become more comfortable with references (and they're very confusing)
It's even more confusing, cause it's not a reference but an alias. Changing elements of @_ means directly changing the passed arguments (see perlsub).
Ironically references shouldn't be confusing, because in most languages arrays and hashes are always copied as references and automatically dereferenced. Perl is one of the rare languages which make a distinction between two flavours.
In JS for instance something like a=;b=a; a=42; b == 42 is true, because a and b are references to the same array.
In Perl you need to explicitly copy references to achieve this, since @b=@a only copies the elements.
But alternatively you can do like JS does with references:
$a=; $b=$a; $a->=42; $b-> == 42
( addicted to the Perl Programming Language)