24 Jan 1998 12:18:47 -0500

Related articles |
---|

LL(2) always factorable to LL(1)? aduncan@cs.ucsb.edu (Andrew M. Duncan) (1998-01-17) |

Re: LL(2) always factorable to LL(1)? clark@quarry.zk3.dec.com (Chris Clark USG) (1998-01-20) |

Re: LL(2) always factorable to LL(1)? bill@amber.ssd.csd.harris.com (1998-01-21) |

Re: LL(2) always factorable to LL(1)? mickunas@cs.uiuc.edu (1998-01-23) |

Re: LL(2) always factorable to LL(1)? cfc@world.std.com (Chris F Clark) (1998-01-23) |

Re: LL(2) always factorable to LL(1)? parrt@magelang.com (Terence Parr) (1998-01-23) |

Re: LL(2) always factorable to LL(1)? thetick@magelang.com (Scott Stanchfield) (1998-01-24) |

Re: LL(2) always factorable to LL(1)? parrt@magelang.com (Terence Parr) (1998-02-01) |

From: | "Scott Stanchfield" <thetick@magelang.com> |

Newsgroups: | comp.compilers |

Date: | 24 Jan 1998 12:18:47 -0500 |

Organization: | MageLang Institute - http://www.MageLang.com |

References: | 98-01-071 98-01-080 98-01-086 |

Keywords: | parse, LL(1) |

Bill Leonard wrote in message 98-01-086...

*>It is true that any LR(k) language is also an LR(1) language, but that*

*>doesn't mean much in relation to LL(k) languages.*

Theory aside for a moment...

Has anyone actually seen an LALR(k) grammar _for_a_useful_language_

that cannot be represented in predicated LL(k)? (Or even in LL(1) if

you left factor like crazy).

Back to theory:

Has anyone proven that the set of languages that can be represented by

a predicated LL(k) grammar is a proper subset of those that can be

represented in LR(k)?

I know (in my gut ;) that it would be at least a much bigger subset, but has

any real thought gone into it?

--

-- Scott

Scott Stanchfield, Santa Cruz, CA

http://www.scruz.net/~thetick

--

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