Re: x + (y) + z
Kelly, >> to >> >> add-expr: >> mul-expr| >> mult-expr '+' add-expr | >> mult-expr '-' add-expr ; >> > >I don't think that's what you want if you're planning >on executing the code you generate from this... A subsequent phase of processing can be used to reverse this syntactic decision. >By the way, you might want to have a look at > >http://www.lysator.liu.se/c/ANSI-C-grammar-y.html Thanks, but this grammar requires a symbol table. It also contains lots of shift/reduce and reduce/reduce conflicts, and if I was writing a more conventional grammar I would want to remove these. derek -- Derek M Jones tel: +44 (0) 1252 520 667 Knowledge Software Ltd mailto:[EMAIL PROTECTED] Applications Standards Conformance Testing http://www.knosof.co.uk ___ Help-bison@gnu.org http://lists.gnu.org/mailman/listinfo/help-bison
Re: x + (y) + z
Frank, >> The statement (y)+z can be parsed as casting >> +z to the type y, or as adding y to z. A couple of >> %dprecs solve this problem (I think the cast is the >> common case for - and a binary expression for +). > >What the "common case" is doesn't really matter since a correct >parser should be able to recognize all valid inputs, not only the >most common ones. So if I understand right what you plan, this would >all be just heuristics which don't really solve the problem. Since both parses will recognise the input one has to be selected. Picking the common case means that there is less work which has to be undone later in the semantics phase. derek -- Derek M Jones tel: +44 (0) 1252 520 667 Knowledge Software Ltd mailto:[EMAIL PROTECTED] Applications Standards Conformance Testing http://www.knosof.co.uk ___ Help-bison@gnu.org http://lists.gnu.org/mailman/listinfo/help-bison
Re: x + (y) + z
At 20:41 + 2005/03/03, Derek M Jones wrote: >The statement (y)+z can be parsed as casting >+z to the type y, or as adding y to z. A couple of >%dprecs solve this problem (I think the cast is the >common case for - and a binary expression for +). The normal way to resolve this would be to let the lexer check the lookup table to see what y is: a type or a number identifier, and then return that type. WHy does this not work for you. >However, things are more complicated for x + (y) + z, >whose parse tree can be either > >+ > /\ > x ( ) > / \ > y + > | > z > >or > > + > / \ >+ z > / \ > x(y) > >As currently implemented the %dprec functionality does >not appear to be any help here. Effectively, it will only >select between two productions that consume the same >number of input tokens. The commands %left and %right handles left and right associativity. Hans Aberg ___ Help-bison@gnu.org http://lists.gnu.org/mailman/listinfo/help-bison
Re: x + (y) + z
Derek M Jones wrote: > >> The statement (y)+z can be parsed as casting > >> +z to the type y, or as adding y to z. A couple of > >> %dprecs solve this problem (I think the cast is the > >> common case for - and a binary expression for +). > > > >What the "common case" is doesn't really matter since a correct > >parser should be able to recognize all valid inputs, not only the > >most common ones. So if I understand right what you plan, this would > >all be just heuristics which don't really solve the problem. > > Since both parses will recognise the input one has to be > selected. Picking the common case means that there is > less work which has to be undone later in the semantics phase. Undoing and reversing parsing (as you also mentioned in another mail) may be an option, but IMHO not a very good one, if you can't use the parser the way it's supposed to. But since you don't state why you don't do the usual thing (separate token for type names), I can't say anything more ... Frank -- Frank Heckenbach, [EMAIL PROTECTED] http://fjf.gnu.de/ GnuPG and PGP keys: http://fjf.gnu.de/plan (7977168E) ___ Help-bison@gnu.org http://lists.gnu.org/mailman/listinfo/help-bison
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