2 May 2005 14:26:03 -0400

Related articles |
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LALR1 and LL1 neelesh.bodas@gmail.com (Neelesh Bodas) (2005-04-11) |

Re: LALR1 and LL1 schmitz@i3s.unice.fr (Sylvain Schmitz) (2005-04-16) |

Re: LALR1 and LL1 148f3wg02@sneakemail.com (Karsten Nyblad) (2005-04-26) |

Re: LALR1 and LL1 schmitz@i3s.unice.fr (Sylvain Schmitz) (2005-04-26) |

Re: LALR1 and LL1 haberg@math.su.se (2005-04-28) |

Re: LALR1 and LL1 148f3wg02@sneakemail.com (Karsten Nyblad) (2005-04-30) |

Re: LALR1 and LL1 schmitz@i3s.unice.fr (Sylvain Schmitz) (2005-05-02) |

Re: LALR1 and LL1 haberg@math.su.se (Hans Aberg) (2005-05-02) |

From: | Sylvain Schmitz <schmitz@i3s.unice.fr> |

Newsgroups: | comp.compilers |

Date: | 2 May 2005 14:26:03 -0400 |

Organization: | Compilers Central |

References: | 05-04-023 05-04-041 05-04-059 05-04-088 |

Keywords: | parse, theory |

Posted-Date: | 02 May 2005 14:26:03 EDT |

Hans Aberg wrote:

*> Do you have any reference? -- Akim Demaille sent me an example where LL(1)*

*> isn't LR(1). :-) I reposted it here, but I have forgotten when.*

I could not find your post in the archives. Anyway this result is a

very sure one; you can find a demonstration for it in _Parsing_Theory_

by Sippu and Soisalon-Soininen, vol. 2, pp. 239--248.

To give a more intuitive answer, an LL(k) parser is like an LR(k) parser

with a semantic action embedded at the very beginning of every single

rule: it has to decide immediately what to do, only looking at the "k"

first tokens from this point of the rule. From there, proper inclusion

of the set of all LL(k) grammars in the set of LR(k) grammars is obvious.

--

Regards,

Sylvain

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