Beefy Boxes and Bandwidth Generously Provided by pair Networks
Perl: the Markov chain saw

Re: Re: Icarus

by larryk (Friar)
on Mar 23, 2001 at 20:42 UTC ( #66664=note: print w/replies, xml ) Need Help??

in reply to Re: Icarus
in thread Icarus

have a look with some formatting...
I'm no expert but I'll have a go at explaining this one.

*$=sub{ $a=q;}die;; sub u(){$_} $= && unpack u=>u }; s[]{82G5S="!A;F]T:&5R(%!E<FP\@:&%C:V5R}six; die &$=>$/

I don't understand the *$ except that * is the typeglob prefix so *test="a" would mean $test is "a" $test[0] (from @test) is "a" and $test{a} is defined but with no value. In this case the Perl processID special variable $$ may be being overwritten with a ref(?) to the sub (someone help me out here please).

The $a line is just to put someone offtrack - the q implies a singlequoted string to follow and the sneaky bit here is that Perl allows a ; to be the delimiter hence the ;;. The offputting bit is that if you put the newline in the wrong place then you get *$=sub{$a=q;} followed by die which appears to make sense. Truth is this whole line can be removed and the output won't be affected.

sub u(){$_} is the same as sub u { return $_; }

$= is a Perl special variable a.k.a. $FORMAT_LINES_PER_PAGE (default 60) and is used here simply as a true value so the unpack will be evaluated.

the s[]{blahblahblah} is a substitution on $_ which is presumably a uuencoded string equating to "Just another Perl hacker" which the unpack reveals (six is just a clever arrangement of unnecessary modifiers to the substitution).

the last line "die &$=>$/" just dies with the returned value from the sub defined earlier although (again help please) the =>$/ (which is the special variable for the input record separator) supresses the "at line X" which usually follows - don't know how though.

how did I do?

Replies are listed 'Best First'.
Re: Re: Re: Icarus
by Adam (Vicar) on Mar 23, 2001 at 23:16 UTC
    Yes, you caught the general ideas, but I'll explain the parts you had questions on. Your formating of the code is correct, and the $a assignment is definatly there to throw you, which it didn't.

    *$=sub{}; is equivalent to &$=sub{}. This is equivalent to sub $ {}, except that you can't do that. The magic is in the typeglob, which recognizes that its getting an anonymous sub, and names it &$. The value of $$ (the PID) is left unchanged. :-)

    Next, you said: sub u(){$_} is the same as sub u { return $_; }, which it isn't. The parens() act as an empty prototype basically telling Perl that the letter u by itself is a valid function call. Otherwise you have to use &u or u(). This allows the u=>u in the next line. Also note that even though this sub is defined inside the scope of the other sub, it is still a package global subroutine - all subroutines are, regardless of where they are defined. Granted there is some magic involved with localized variables and closures, but this doesn't run into any of that.

    Is kind of cool in that it takes $_, which is undefined, and replaces it with the payload, uuencoded. The six are just useless extensions, but distracting ones. Its also sort of a joke in that this was my sixth obfuscation posted here. Note that this will generate a warning about using an undefined variable, but will still pass strict.

    And last but not least, die &$=>$/ calls the sub defined earlier, which basically does die unpack( u=>u), $/ Now you asked about $/, it's default value is "\n". If you die with a string ending in a newline, then the "at line X" is supressed. That's also true with warn. The u=>u is equivalent to "u", u() where the first u tells unpack to uudecode the return from u(), which just returns $_.

    Good job of taking this obfu apart.


      I have just one other query. You seem to have used => as a comma replacement. I know this is alternate notation when defining a hash but will Perl always interpret this as a comma?

        From perlop

        The => digraph is mostly just a synonym for the comma operator. It's useful for documenting arguments that come in pairs. As of release 5.001, it also forces any word to the left of it to be interpreted as a string.

Re: Re: Re: Icarus
by jorg (Friar) on Mar 23, 2001 at 22:00 UTC
    aah yes indeed i'm starting to get the picture now thanks larryk, ++ tomorrow 'cuz haven't got any votes left.
    thanks again!


    "Do or do not, there is no try" -- Yoda

Log In?

What's my password?
Create A New User
Node Status?
node history
Node Type: note [id://66664]
and all is quiet...

How do I use this? | Other CB clients
Other Users?
Others wandering the Monastery: (2)
As of 2017-03-28 03:18 GMT
Find Nodes?
    Voting Booth?
    Should Pluto Get Its Planethood Back?

    Results (326 votes). Check out past polls.