hen a (sub)parser fails to match the input, and it will always backtrack 
        to the next choice point upward (not backward) in the parser structure. 
        In other words abb|ab will match &quot;ab&quot;, as will a(bb|b), but 
        (ab|a)b won't because the (ab|a) subparser will always match the 'b' after 
        the 'a' if it is available.</p>
      <p>--Rainer Deyke</p>
      </td>
  </tr>
</table>
<p>There's a strong preference on &quot;simplicity with all the knobs when you 
  need them&quot;