|Is LL(k) LL(1) ? feedMEemail@example.com (1998-04-15)|
|Re: Is LL(k) LL(1) ? firstname.lastname@example.org (William D Clinger) (1998-04-29)|
|Re: Is LL(k) LL(1) ? corbett@lupa.Eng.Sun.COM (1998-05-04)|
|Re: Is LL(k) LL(1) ? email@example.com (Torben Mogensen) (1998-05-07)|
|Re: Is LL(k) LL(1) ? firstname.lastname@example.org (Joseph H. Fasel) (1998-05-12)|
|From:||"Joseph H. Fasel" <email@example.com>|
|Date:||12 May 1998 22:24:08 -0400|
|Organization:||Los Alamos National Laboratory|
|References:||98-04-065 98-04-107 <98-05-013@com 98-05-029|
corbett@lupa.Eng.Sun.COM (Robert Corbett) writes:
> >Any programming language that contains the dangling-else construct
> >is not LL(k) for any k.
Torben Mogensen wrote:
> This is only half right. No grammar describing dangling else is LL(k),
> but it is easy to construct an LL(1) parse table that handles the
> dangling else problem. Hence, the language is LL(1) while the grammar
> is not.
I thought the definition of an LL(k) language was one that has an
LL(k) grammar. Is there some other definition in terms of parsers?
Joseph H. Fasel, Ph.D. email: firstname.lastname@example.org
Technology Modeling and Analysis phone: +1 505 667 7158
University of California fax: +1 505 667 2960
Los Alamos National Laboratory post: TSA-7 MS F609; Los Alamos, NM 87545
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