Not what you'd expect

It's fun to use obfu to work on your regex skills...

#!/usr/bin/perl
use strict;
$_= "Just Another Perl Hacker";
print /(.)n/;
s/^([^ ]*) *([^ ]*)/$3 $3/;
/\sH(.)(.)/;
print $2;/(.)\1.(.)/;
print $2,",";print $_;
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Comment on Not what you'd expect Download Code

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Solution: Not what you'd expect by domm (Chaplain) on Mar 09, 2002 at 19:40 UTC
OK, let's de-obfuscate... $_= "Just Another Perl Hacker"; print /(.)n/; That's easey, set $_ to JAPH, and print the first char followed by an 'n', i.e. 'A'. `s/^([^ ]) ([^ ])/$3 $3/;` Here you have two capturing parenthenses, but replace with $3, which is undef. So in fact you are removing something of $_. And what are you removing? `^([^ ])` Thats, from the beginning of the string, everything that's not a space, until the next space, i.e. 'Just'. `([^ ])` This is basically the same regex, removing the next word (i.e. 'Another') `/\sH(.)(.)/;` This is no replacement, but a match for the two characters following ' H', that is 'a' and 'c'. `print $2;` Print the second match, i.e. 'c'. `/(.)\1.(.)/;` Another match, the most tricky in this obuf (IMO). Reminder: \1 in a regex is basically the same as $1 outside, i.e. the value found in the first capturing parenthenses. So, this matches the first char followed by itself, and saves the char after the next in $2. AS you replaced the first two words with nothing, $_ now looks like (using _ instead of space for clarity) `__Perl Hacker`, so the regex matches __Perl Hacker and thus saves 'e' in $2; `print $2,",";` Which you print here, followed by a ','. `print $_;` Print $_ (BTW, just `print` would've been enough..), .i.e. ' Perl Hacker' Not that hard to de-obfuscate, the only thing that made me wonder what was going on was '\1' in one regex. -- #!/usr/bin/perl -w just another perl hacker print+seek(DATA,$=*.3,@-)?~~<DATA>:$:__DATA__	[reply] [d/l] [select]

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Solution: Not what you'd expect
by domm (Chaplain) on Mar 09, 2002 at 19:40 UTC

$_= "Just Another Perl Hacker";
print /(.)n/;

s/^([^ ]*) *([^ ]*)/$3 $3/;
Here you have two capturing parenthenses, but replace with $3, which is undef. So in fact you are removing something of $_. And what are you removing?
^([^ ]*)
Thats, from the beginning of the string, everything that's not a space, until the next space, i.e. 'Just'.
*([^ ]*)
This is basically the same regex, removing the next word (i.e. 'Another')

/\sH(.)(.)/;
This is no replacement, but a match for the two characters following ' H', that is 'a' and 'c'.

print $2;
Print the second match, i.e. 'c'.

/(.)\1.(.)/;
Another match, the most tricky in this obuf (IMO).
Reminder: \1 in a regex is basically the same as $1 outside, i.e. the value found in the first capturing parenthenses.
So, this matches the first char followed by itself, and saves the char after the next in $2. AS you replaced the first two words with nothing, $_ now looks like (using _ instead of space for clarity) __Perl Hacker, so the regex matches

__Perl Hacker

print $2,",";
Which you print here, followed by a ','.

print $_;
Print $_ (BTW, just print would've been enough..), .i.e. ' Perl Hacker'

Not that hard to de-obfuscate, the only thing that made me wonder what was going on was '\1' in one regex.

--
#!/usr/bin/perl -w just another perl hacker
print+seek(DATA,$=*.3,@-)?~~<DATA>:$:__DATA__

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