On 26 May 2017 at 05:33, lee <l...@yagibdah.de> wrote: > Perl doesn't have data structures, so variables in perl are not data > structures. That is unfortunate.
So when I write: my $var = { memory => { total => 1024, free => 100, buffers => 10, }, }; What do I have? Because as far as I'm concerned, I have both data structures and references in play. References are not a low level mechanic that only the Perl VM needs to care about. References are a mechansim to allow data structures to be passed *without copying* my $x = 5; my $y = $x; # x is a copy of y However, if I do: my $x = 5; $y = \$x Y is now a reference to X If I now do: ${$y} = 10 X changes. Its a useful tool, that programmers have uses for. If you think they're evil, then you're probably thinking too much in C. Because you can't do any real work in perl *without* references. *Objects* in perl are blessed references. Avoid that all you like, but you're just avoiding using Perl and wondering why it hurts :) -- Kent KENTNL - https://metacpan.org/author/KENTNL -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/