On Thu, Aug 31, 2000 at 09:46:54PM +0200, Fredrik Lundh wrote:
can anyone tell me how Perl treats this pattern? r'((((((((((a))))))))))\41'
if I understand this correctly, Perl treats as an *octal* escape (chr(041) == "!").
Correct. From perlre: You may have as many parentheses as you wish. If you have more than 9 substrings, the variables $10, $11, ... refer to the corresponding substring. Within the pattern, \10, \11, etc. refer back to substrings if there have been at least that many left parentheses before the backreference. Otherwise (for backward compatibility) \10 is the same as \010, a backspace, and \11 the same as \011, a tab. And so on. (\1 through \9 are always backreferences.) In other words, if there were 41 groups, \41 would be a backref to group 41; if there aren't, it's an octal escape. This magical behaviour was deemed not Pythonic, so pre uses a different rule: it's always a character inside a character class ([\41] isn't a syntax error), and outside a character class it's a character if there are exactly 3 octal digits; otherwise it's a backref. So \41 is a backref to group 41, but \041 is the literal character ASCII 33. --amk