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World: r3wp

[Parse] Discussion of PARSE dialect

Steeve
28-Jun-2007
[2109]
we want any combination of "ABC"
Tomc
28-Jun-2007
[2110]
without repeats
Steeve
28-Jun-2007
[2111x2]
yep
and that works for any type of sub-rule, no only for constants
Tomc
28-Jun-2007
[2113x2]
one shot concept
>> data: "abc"
== "abc"

>> c: charset "abc" parse data [3 [here:  c  (c: difference c to 
bitset! first :here)]]
== true
>> data: "ccc"
== "ccc"

>> c: charset "abc" parse data [3 [here:  c  (c: difference c to 
bitset! first :here)]]
== false
>>
Steeve
28-Jun-2007
[2115]
hehe, doen't work if i check for any combinations of "AABC"
Tomc
28-Jun-2007
[2116]
I should read the thread  when I have a few minutes
Steeve
28-Jun-2007
[2117]
yes you should
Tomc
28-Jun-2007
[2118]
but I will amuses myself as I please in the meanwhile.
Steeve
28-Jun-2007
[2119]
huhu
Brock
28-Jun-2007
[2120x2]
Steve, wondering if your take function solution was coded in french, 
I can't understand it  without studying it a bit more.
;-)
Steeve
28-Jun-2007
[2122x2]
it's the french touch ;-)
hey Brock , still alive ?
Tomc
29-Jun-2007
[2124x2]
do rules: [
	set 'a "first rule "
	set 'b "second rule "
	set 'c "third rule "
]

rule: [ 
	[end (if empty? erode[here: back tail :here]) :here skip]| 
	[here: erode there: 
	(	all[
			token: copy/part :here :there
			word: find rules token
			word: first back word
			remove/part find erode word 2 
		]

  all[empty? erode not tail? there insert/only erode [] there: tail 
  there]
	)] :there
    rule
]

data: "first rule second rule third rule "
erode: [ a | b | c | a]  parse/all data rule
left expanded for clarity
Steeve
30-Jun-2007
[2126x5]
i would have to discuss what follow before posting it in Rambo
parse "  a" [copy r "a"]
>>r
== " a"
i think it should be r = "a" instead
*i would have speech
Tomc
30-Jun-2007
[2131x5]
>> parse/all " a" [any " " copy r "a"]
== true
>> r
== "a"
I try toi allways yse /all when I give a block of rules
and use pases by itself aith the none rule
r: parse " a" none
["a"]
Steeve
30-Jun-2007
[2136x3]
hi Toms, as always don't focus on the example i give but on the issue 
i suggest, another one example:
>>parse "ld   a  b" ["ld" "a" copy reg ["b" | "c"]]
>> r == " b" 
>> i just think it will be more logic to have r == "b"
(replace r by reg)
i well know how to skip blank chars, but i think it's not logic in 
that case
[unknown: 9]
30-Jun-2007
[2139]
We so need a Wiki for Rebol, and shove every word and example into 
it.  Just Parse needs 100 pages of examples and descriptions.
ICarii
30-Jun-2007
[2140]
lol
Steeve
30-Jun-2007
[2141]
we need a wiki just for parse
ICarii
30-Jun-2007
[2142]
and a new book.. Dialects for The Rest of Us (forget the Dummies 
version..)
[unknown: 9]
30-Jun-2007
[2143]
Exactly!
Steeve
30-Jun-2007
[2144]
books books, scripts in rebol.org are enough for me
[unknown: 9]
30-Jun-2007
[2145]
There is always another trick, and another level to Rebol, the docs 
NEED to be in a Wiki, where we can add to them, let them live, breath...
ICarii
30-Jun-2007
[2146]
there is a rebol wikibooks
[unknown: 9]
30-Jun-2007
[2147]
Yeah, it needs someone to kick it off, get it started, fill it with 
about 50 good words.  Where is a "kid" when you need one?
ICarii
30-Jun-2007
[2148]
playing with their new iphones and parsing text messages internally 
:)
Steeve
30-Jun-2007
[2149x3]
so, spending money to be cool
(but what is the interest to purchase what everyone wants, that's 
not cool at all)
REBOL is cool, and free
Tomc
1-Jul-2007
[2152x2]
Steeve do you have any case where the self modifying recursive parse 
rule above fails?
>> ; insufficent pattern
>> data: "first rule second rule third "
== "first rule second rule third "
>> erode: [a | b | c]
== [a | b | c]
>> parse/all data rule
== false
>>
>> ;;; unused rule before pattern is consumed
>> data: "first rule second rule third rule "
== "first rule second rule third rule "
>> erode: [a | b | c | a]
== [a | b | c | a]
>> parse/all data rule
== false
>>
>> ;;; patterns and rules allowed in any order
>> data: "first rule second rule third rule "
== "first rule second rule third rule "
>> erode: [c | b | a]
== [c | b | a]
>> parse/all data rule
== true
>>
>> ;;; multiple (duplicate) rules allowed
>> data: "first rule first rule second rule third rule "
== "first rule first rule second rule third rule "
>> erode: [a | b | c | a]
== [a | b | c | a]
>> parse/all data rule
== true
>>
Steeve
1-Jul-2007
[2154x2]
no i don't, it's an intersting alternative to my own code
a crazy simple alternative from DocKimbel (non repetitive pattern 
and limited to rules with single char)

>> rule: [(c: charset "ABC") 3 [copy v c (remove/part c v)]]

>> parse "ABC" rule
== true
>> parse "BAC" rule
== true
>> parse "CBA" rule
== true
>> parse "ABA" rule
== false
>> parse "ABCA" rule
== false
btiffin
6-Jul-2007
[2156]
How do you build parse rules?

rule: copy []
word: "abc"

;; Want to compose the block to look like this  ["abc" (print ["found 
word len" 3])] 

insert tail rule compose [(word) (print "found word len (length? 
word))]  no go - obvious

I've tried  compose/deep [(word) (to paren! [print ["found word len" 
(length? word)])]  but length? word doesn't get composed, it gets 
included in the to paren! expression  compose/only/deep same thing


I guess the question is what is the technique to compose something 
that is to include paren! expressions with data inside that needs 
to be composed?  **With binding of course**  :)
Chris
7-Jul-2007
[2157x2]
reduce [word to paren! compose/deep [print ["Found Word of Length" 
(length? word)]]]
compose [(word) (to paren! compose/deep [print ["found word len" 
(length? word)]])]