- From: Jia Pu <JiaPu@LumenVox.com>
- Date: Thu, 8 Apr 2004 13:41:35 -0700
- To: <www-voice@w3.org>
I have question on example 17 in the draft.
$x = $a<1-> {$$};
$$ resolves to the last occurrence of $a.
The effect is equivalent to $x = ($a {$$})<1->.
I think there is one difference between the two rules. In the first one,
{$$} only gets evaluated once when $x was matched. But in the second case,
($$) gets evaluated every time $a was matched.
If we parse "111" using following grammars:
1.
root $x;
$a = '1';
$x = {$=""} $a<1-> {$+=$$};
2.
root $x;
$a = '1';
$x = {$=""} ($a{$+=$$})<1-> ;
I think the first one gives us $x="1", and the second one $x="111".
Am I right?
Jia (Jay) Pu
Software Engineer
LumenVox LLC
-------------------------------------------
Anytime a linguist leaves the group
the recognition rate goes up.
--Fred Jelinek
(then of the IBM speech group) (1988)
Received on Thursday, 8 April 2004 16:44:20 UTC