$_= "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 Hackerand 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__
In reply to Solution: Not what you'd expect
by domm
in thread Not what you'd expect
by ProgrammingAce
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