On Fri, May 07, 2004 at 12:31:28PM -0700, Jorge Llamb?as wrote:
>
> --- Robin Lee Powell <rlpowell@digitalkingdom.org> wrote:
> > > What does {da zo si si} do?
> >
> > It *should* result in just 'da', because zo is defined as turning
> > itself and the next argument into a single word. zoi is *not* so
> > defined.
>
> Is there a justification for that difference?
Huh.
I'm sorry, I thought that that difference was explicitely stated
somewhere, but I can't find it. The Red Book doesn't seem to say
whether or not "zo da" is treated as a single word at all.
So unless I'm missing something, all we have to go on is grammar.300,
which says:
a. If the Lojban word "zoi" (selma'o ZOI) is identified, take the
following Lojban word (which should be end delimited with a pause for
separation from the following non-Lojban text) as an opening delimiter.
Treat all text following that delimiter, until that delimiter recurs
*after a pause*, as grammatically a single token (labelled
'anything_699' in this grammar). There is no need for processing within
this text except as necessary to find the closing delimiter.
b. If the Lojban word "zo" (selma'o ZO) is identified, treat the
following Lojban word as a token labelled 'any_word_698', instead of
lexing it by its normal grammatical function.
So "zoi da de da" is turned into four tokens, "zoi da anything_699 da"
and "zo da" is turned into the single token "any_word_698".
It is this behaviour that I am trying to emulate, without being a slave
to YACC restrictions (which, for example, make it so that "zoi da weeble
da si si si si" works, but no lesser number of "si" after the zoi have
any effect.