This is as far as I got:
^(?=.{65}\\\\Q\\1\\\\E\\\)\z)(\Q^(?=.{65}\\\\Q\\1\\\\E\\\)\z)(\E)\\Q\1\\E\)
It is a true inequ of sorts. However, it uses \Q, which is a quoting
construct, not a regex construct. So it doesn't work as $x =~ /$x/,
only as $x =~ eval("qr/$x/"):
chomp(my $x = <<'EOF');
^(?=.{65}\\\\Q\\1\\\\E\\\)\z)(\Q^(?=.{65}\\\\Q\\1\\\\E\\\)\z)(\E)\\Q\1\\E\)
EOF
print "OK" if $x =~ eval("qr/$x/");
Going all the way to $x = qr/.../; "$x" =~ $x; doesn't help either,
because in the stringyfied version \Q-quoting is replaced by normal
\-escaping.
- Karsten
