Hi,

In message "[ruby-talk:15938] Re: Regexp (a\1)"
    on 01/05/30, ts <decoux / moulon.inra.fr> writes:

|  "aaaaaa"=~ /^(a\1?){4}$/
|
| must give $1 = "aa"

Thank you.  I confirmed your fix in [ruby-talk:15924] works like Perl.
But I'm still not sure which is more intuitive.  Although probably I
should stick to Perl's behavior.

							matz.