Related articles |
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[2 earlier articles] |
Re: Regular expression grammar? terryg@uswest.net (1999-09-20) |
Re: Regular expression grammar? lex@cc.gatech.edu (1999-09-20) |
Re: Regular expression grammar? cbarron3@ix.netcom.com (1999-09-20) |
Re: Regular expression grammar? zalman@netcom18.netcom.com (Zalman Stern) (1999-09-24) |
Re: Regular expression grammar? zalman@netcom18.netcom.com (Zalman Stern) (1999-09-24) |
Re: Regular expression grammar? zalman@netcom18.netcom.com (Zalman Stern) (1999-09-24) |
Re: Regular expression grammar? cbarron3@ix.netcom.com (1999-09-27) |
From: | cbarron3@ix.netcom.com (Carl Barron) |
Newsgroups: | comp.compilers |
Date: | 27 Sep 1999 11:18:54 -0400 |
Organization: | Netcom |
References: | 99-09-051 99-09-087 |
Keywords: | lex |
Zalman Stern <zalman@netcom18.netcom.com> wrote:
> I sent email to the original poster 'cause I was embarassed to post my
> grammar, but seeing the responses, I think its ok. My main point to the
> poster is that the grammar needs to tell the parser generator how to handle
> the following two cases:
> is "ab|cd" equivalent to "a(b|c)d" or equiv. to "(ab)|(cd)"
> and
> is "ab*" equivalent to "a(b*)" or equiv. to "(ab)*"
ab}cd is normally (ab)|(cd)
ab* is normally a(b*)
and for all concerned a** == a*
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