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| Related articles |
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| Grammar needed leonardo@dcc.ufmg.br (Leonardo Teixeira Passos) (2006-10-24) |
| Re: Grammar needed schmitz@i3s.unice.fr (Sylvain Schmitz) (2006-10-26) |
| Re: Grammar needed 148f3wg02@sneakemail.com (Karsten Nyblad) (2006-10-26) |
| Re: Grammar needed pjj@cs.man.ac.uk (Pete Jinks) (2006-10-26) |
| Re: Grammar needed cfc@shell01.TheWorld.com (Chris F Clark) (2006-10-26) |
| Re: Grammar needed leonardo@dcc.ufmg.br (Leonardo Teixeira Passos) (2006-11-01) |
| From: | Sylvain Schmitz <schmitz@i3s.unice.fr> |
| Newsgroups: | comp.compilers |
| Date: | 26 Oct 2006 00:29:34 -0400 |
| Organization: | Compilers Central |
| References: | 06-10-101 |
| Keywords: | parse, theory |
| Posted-Date: | 26 Oct 2006 00:29:34 EDT |
| X-Virus-Scanned: | amavisd-new at i3s.unice.fr |
Leonardo Teixeira Passos wrote:
> I've been trying to obtain a grammar that meets two properties:
>
> (i) It represents an LR(k) language
> (ii) The grammar it self isn't LR(k) for any k, for there is at least
> one conflict. One or more of these conflicts does not indicate
> ambiguity in the grammar, but can't be solved with any k.
You can consider for instance the grammar with rules
S -> A a | B b
A -> A c | c (G1)
B -> B c | c
(i) The language generated by this grammar is a deterministic language
`c+(a|b)', since it is also generated by the LR(0) grammar G2 with rules
S -> C a | C b (G2)
C -> C c | c
Or if you want to keep the two parts separated, by the SLR(1) grammar G3
with rules
S -> A a | B b
A -> c A | c (G3)
B -> c B | c
(ii) The grammar G1 is not LR(k) for any k but unambiguous: in order to
choose between the reduction of the first `c' to `A' or `B', the parser
needs the see the ending `a' or `b', which can be arbitrarily far away
after a sequence of `c's.
--
Hope that helps,
Sylvain
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