```\$;                              # A lone dollar?
=\$";                            # Pod?
\$;                              # The return of the lone dollar?
{Just=>another=>Perl=>Hacker=>} # Bare block?
=\$/;                            # More pod?
print%;                         # No right operand for %?

-- Abigail

Comment on Things are not what they seem like.
RE: Things are not what they seem like.
by Russ (Deacon) on Jul 13, 2000 at 11:01 UTC
Okay, now this is just entirely too cool!

I especially love the use of ; in new and previously unknown (for me) ways.

Wow!

Russ
Brainbench 'Most Valuable Professional' for Perl

RE: Things are not what they seem like.
by princepawn (Parson) on Jul 25, 2000 at 00:04 UTC
I wouldlike some explanation of this code... you are clearly thundering from the great heights with this code, but w/out explanation, a poor sap like me is left in the dark.
Okay, here's how it works.

Let's first clean it up (which takes all the fun out of it, but still...):

```    \$; = \$";
\$;{Just=>another=>Perl=>Hacker=>} = \$/;
print %;
Looks more understandable, now, right?

The first line sets the value of the special variable \$; to the value of \$". \$" is the list separator and has the default value of a space. \$; is the subscript separator, which is used (or used to be used) for multidimensional array emulation. As explained in perlvar. So saying

```    \$foo{\$a, \$b, \$c}
really means
```    \$foo{ join \$;, \$a, \$b, \$c }
Since we've set \$; equal to the value of \$", the subscript separator is now a space (' ').

Next line, then:

```    \$;{Just=>another=>Perl=>Hacker=>} = \$/;
Let's fix it up a bit:
```    \$;{Just,another,Perl,Hacker,} = \$/;
That actually isn't legal, though, because the special => makes it okay to use the barewords. If we replace them with commas, we'll get errors. And that's why we need the => after "Hacker"; if we take it off, we get an error.

Anway, though, now it makes more sense, doesn't it? Because it looks like the example above, the example from perlvar. We're just assigning to a hash element in the hash %;.

And \$/ is the input record separator, the default of which is a carriage return ("\n"). So we assign that value to the hash element, so what we really have is something like this:

```    \$;{ join ' ', "Just", "another", "Perl", "Hacker", "" }
= "\n";
Which is just this:
```    \$;{"Just another Perl Hacker"} = "\n";
And then we're at the last line:
```    print %;
Which is very simple. We're just printing out the hash %;, which we just assigned to. In list context, the hash is flattened to a list. This list, in fact:
```    ("Just another Perl Hacker", "\n")
And what happens when we print out that list? Just what you'd expect:
```    Just another Perl Hacker
So that's it. Doesn't it make you love Perl? :)

# I cannot + vote you enough on this one

To some, Perl is an art form. To others it is a religion. What is it to you?
Excelent use of special varibles, and excelent explanation!
oh wow! This is Brilliant, Awesome stuff, Great explanation. Thanks a ton, and Oh Yeah it makes me go mad about Perl!
(zdog) RE: Things are not what they seem like.
by zdog (Priest) on Aug 01, 2000 at 06:19 UTC
Great post!

At first I thought, what the...? But then I realized that \$; was the name of a variable. Next I say, "AHA! Does it work with use strict ?" At first I didn't think so since I thought that you would have to declare \$; by saying "my \$;", but then I hopped on over to perlvar and realized that \$; is a "special variable" and that set me straight.

And a few moments later, I said, "Doh! I should have just read btrott's post and it would have saved me a lot of trouble. Oh, well."

Zenon Zabinski | zdog | zdog7@hotmail.com

Re: Things are not what they seem like.
by Lucky (Scribe) on Dec 05, 2001 at 16:05 UTC
It's probably the most beatiful things I saw. Great!!!
Re: Things are not what they seem like.
by manav (Scribe) on Mar 03, 2005 at 11:17 UTC
Copy/Paste this code into a script(say wow.pl) and then
perl -MO=Deparse wow.pl

Manav