six lol. very good - i like it.
it's a bit over my head but I get the general idea. | [reply] |
would you care to explain this one to me? As obfuscation goes completely over my head yours looks like a nice and short example that i *might* be able to comprehend
Jorg
"Do or do not, there is no try" -- Yoda | [reply] |
Here it is formated nicly:
*$ = sub{
$a=q;}die;;
sub u{
$_;
}
$= && unpack u => u;
};
s[]{82G5S="!A;F]T:&5R(%!E<FP\@:&%C:V5R}six;
die &$ => $/;
Boiled down this can be represented by
*$=sub{
unpack "u", $_;
};
$_='82G5S="!A;F]T:&5R(%!E<FP@:&%C:V5R';
die &$, $/;
or even a one-liner:
die unpack ("u", '82G5S="!A;F]T:&5R(%!E<FP@:&%C:V5R'), $/;
With this, you should be able to figure out what was done to hide the code. | [reply]
[d/l]
[select] |
have a look with some formatting...
I'm no expert but I'll have a go at explaining this one.
*$=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 $$ may be being overwritten with a ref(?) to the sub (someone help me out here please).
The $a line is just to put someone offtrack - the q implies a singlequoted string to follow and the sneaky bit here is that Perl allows a ; to be the delimiter hence the ;;. The offputting bit is that if you put the newline in the wrong place then you get *$=sub{$a=q;} followed by die which appears to make sense. Truth is this whole line can be removed and the output won't be affected.
sub u(){$_} is the same as sub u { return $_; }
$= is a Perl special variable a.k.a. $FORMAT_LINES_PER_PAGE (default 60) and is used here simply as a true value so the unpack will be evaluated.
the s[]{blahblahblah} is a substitution on $_ which is presumably a uuencoded string equating to "Just another Perl hacker" which the unpack reveals (six is just a clever arrangement of unnecessary modifiers to the substitution).
the last line "die &$=>$/" just dies with the returned value from the sub defined earlier although (again help please) the =>$/ (which is the special variable for the input record separator) supresses the "at blah.pl line X" which usually follows - don't know how though.
how did I do? | [reply]
[d/l] |
Yes, you caught the general ideas, but I'll explain the parts you had questions on. Your formating of the code is correct, and the $a assignment is definatly there to throw you, which it didn't.
*$=sub{}; is equivalent to &$=sub{}. This is equivalent to sub $ {}, except that you can't do that. The magic is in the typeglob, which recognizes that its getting an anonymous sub, and names it &$. The value of $$ (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.
s[]{82G5S="!A;F]T:&5R(%!E<FP\@:&%C:V5R}six;
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 blah.pl 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.
| [reply]
[d/l] |
aah yes indeed i'm starting to get the picture now thanks larryk, ++ tomorrow 'cuz haven't got any votes left.
thanks again!
Jorg
"Do or do not, there is no try" -- Yoda
| [reply] |
