[REBOL] Re: Parse versus Regular Expressions
From: andreas:bolka:gmx at: 3-Apr-2003 10:56
Thursday, April 3, 2003, 9:39:43 AM, Ladislav wrote:
> Then he asked me, whether the following equation could be easily
> rewritten for PARSE too:
> Y = Y.a + b
> The result of the equation is known, it is:
> Y = ba*
> , i.e. in PARSE notation:
> y: [#"b" any #"a"]
> Noticeably, the "dumb" approach to the equation rewriting:
> y: [[y #"a"] | #"b"]
I just want to add, that this is not necessarily a "dump" approach but
rather a left-recursive approach. Wether a given parser can handle
left-recursive grammars depends on the parser's internal
implementation.
There are certainly situations where left-recursive grammars are by
far more natural and elegant when compared to their non left-recursive
aequivalents.
> didn't "fire" any error, stack overflow, or any unusual exception.
> Nevertheless, it didn't work:
> parse "b" y ; == true
> parse "ba" y ; == false
> Any ideas?
Not really, sorry (unless you count my conclusion below as "ideas" :).
The problem here is that we (at least I :) do not know how 'parse is
implemented internally (and I have not attempted to gain more
knowledge by experimentation).
The only things I dare to conclude from my experiences (or from your
examples):
- 'parse is not able to handle left-recursive grammars
- 'parse does not tell you, when it encounters left recursion, it
"silently" pretends that the sentence (the "ba" in your example)
does not match the grammar. Not a very desireable behaviour, imho.
--
Best regards,
Andreas mailto:[andreas--bolka--gmx--net]
