## [REBOL] Parse versus Regular Expressions

### From: lmecir:mbox:vol:cz at: 3-Apr-2003 9:39

Hi all,
sorry, if this is too trivial for the Regular Expressions (RE) experts, I am certainly
not one (I am not using RE at all, neither I knew the definition of RE).
I've heard, that PARSE was more general, than RE, although I didn't know why. Only recently
I found an example:
anbn: [#"a" anbn #"b" | none]
parse "aaaaabbbbb" anbn ; == true
The following is an RE equation ("." means concatenation and "+" means something like
or):
X = a.X + b
, which can be trivially rewritten for PARSE as follows:
x: [[#"a" x] | #"b"]
and the result:
parse "b" x ; == true
parse "ab" x ; == true
parse "aaab" x ; == true
Of course, the result of the RE equation is known, it is ("*" means repetition):
X = a*b
, which can be rewritten for PARSE as:
x: [any #"a" #"b"]
, but the fact, that both approaches worked has been appreciated by a local teacher of
the subject. 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"]
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?
Regards
-L