Related articles |
---|
Grammar -> Parser question chris63@my-dejanews.com (1998-06-04) |
Re: Grammar -> Parser question td@sysinno.se (Torbjorn Drevin) (1998-06-09) |
Re: Grammar -> Parser question torbenm@diku.dk (Torben Mogensen) (1998-06-09) |
Re: Grammar -> Parser question dwight@pentasoft.com (1998-06-09) |
Re: Grammar -> Parser question qjackson@wave.home.com (Quinn Tyler Jackson) (1998-06-09) |
Re: Grammar -> Parser question qjackson@wave.home.com (Quinn Tyler Jackson) (1998-06-18) |
From: | Torben Mogensen <torbenm@diku.dk> |
Newsgroups: | comp.compilers |
Date: | 9 Jun 1998 11:59:44 -0400 |
Organization: | Compilers Central |
References: | 98-06-018 98-06-029 |
Keywords: | parse, LL(1) |
Chris writes:
>Suppose I have the following Grammar:
>condition -> expr PLUS expr |
> LPAREN condition RPAREN
>
>expr -> ID |
> LPAREN expr RPAREN
>I want to hand-code a top down recursive descent parser for this grammar
>using ideas from the Dragon Book (nonterminals become functions, terminals
>become calls to match()). I can't use lex or yacc.
>However, the LPAREN symbol is causing a problem: When I see it, I don't
>know if I'm dealing with a nested condition or a nested expression.
This example AFAICS can't be hadled by finite nfolding and left
factoring alone, which is otherwise often the case with similar
examples. In cases such as the above, a typical solution is to make a
nonterminal that recognizes (a superset of) the union of expr and
condition and then filter out the bad ones by using attributes.
The combined nonterminal has the productions
conexp -> ID
| conexp PLUS expr
| LPAREN conexp RPAREN
Ee eliminate the left recursion in the usual way:
conexp -> ID conexp'
| LPAREN conexp RPAREN conexp'
conexp' -> PLUS expr conexp'
| \epsilon
We then use a synthesized attribute which tells us if the result is a
condition or an expr. Using notation similar to the Dragon Book, we
can write the attributed grammar as
conexp -> ID conexp' { conexp.isexpr := conexp'.isempty }
conexp -> LPAREN conexp1 RPAREN conexp'
{ if conexp1.isexpr
then conexp.isexpr := conexp'.isempty
else if conexp'.isempty
then conexp.isexpr := false
else error }
conexp' -> PLUS expr conexp'1
{ conexp'.isempty := false }
conexp' -> \epsilon { conexp'.isempty := true }
Torben Mogensen (torbenm@diku.dk)
--
Return to the
comp.compilers page.
Search the
comp.compilers archives again.