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World: r3wp

[Parse] Discussion of PARSE dialect

btiffin
27-Jun-2007
[1951]
Yep.  I'm out.  Not experienced enough to have piped up...  ;)
Steeve
27-Jun-2007
[1952x2]
and what is your mater ? ;-)
we can talk of rebcode if you want :-)
btiffin
27-Jun-2007
[1954]
Thinking about this one, you'd have to repeat all the alternatives. 
 Defeating the purpose...but I'll bet Chris or Geomol or Anton or... 
will have a workable solution posted before ya know it.
Steeve
27-Jun-2007
[1955]
i h'ave many alternatives (all comsuming lot of code)  that's not 
the point, i just think it could be more elegant to have such a command
btiffin
27-Jun-2007
[1956]
Maybe we could hold side bets on which rebol will come up with the 
solution first...I'd put money down on Oldes or Volker giving you 
a workable  :)
Steeve
27-Jun-2007
[1957]
if you say so, i  tried many alternatives, but they are all uggly
Brock
27-Jun-2007
[1958]
try this... parse "AB" [some ["A" |  "B" | "C"] | none 
]
Steeve
27-Jun-2007
[1959]
i don't want to parse "AB" but any combinations of "ABC"
btiffin
27-Jun-2007
[1960]
His works.  I tried it without the terminating  | none  and it spun 
out, so I didn't mention it  :)
Steeve
27-Jun-2007
[1961x2]
parse "AB" sould return false
not true
Brock
27-Jun-2007
[1963x2]
why would AB return false?
so it must be three characters?
Steeve
27-Jun-2007
[1965x2]
because i want parse the complete serie "ABC" in any combination
yes 3 characters
btiffin
27-Jun-2007
[1967]
What it you put 1 repeats inside the some?
Steeve
27-Jun-2007
[1968]
write it plz
btiffin
27-Jun-2007
[1969]
nevermind...  just tried it.   piping up out of turn again  :)
Steeve
27-Jun-2007
[1970x2]
i just don't understand what you mean
ah ok, it doesn't work :-)
Brock
27-Jun-2007
[1972]
do all three characters need to be present?
btiffin
27-Jun-2007
[1973]
I tried  parse "AB" [some [1 "A" | 1 "B" | 1 "C"] | none]   ...returns 
true
Brock
27-Jun-2007
[1974]
or can characters be duplicated?
Steeve
27-Jun-2007
[1975]
no Brock
Brock
27-Jun-2007
[1976]
the | none handles situations where the rule doesn't apply, so almost 
everything will return true
btiffin
27-Jun-2007
[1977]
Brock;  Without it it'll endless loop no?
Steeve
27-Jun-2007
[1978x2]
no
just never use an ''opt" rule in a some or any block
Brock
27-Jun-2007
[1980]
will it just be instances of A B & C?
Steeve
27-Jun-2007
[1981x6]
one instance of each in any order (but just one instance)
the awaiting result is 
a: b: c: false

parse "CAB" [some ["A" (a: true) | "B" (b: true) | "C" (c: true)]] 
if all [a b c ] [print "ok"]
*expecting result
except that this code is wrong for series with repeting characters 
like "AABCC"
(should increment a b and c instead using boolean)
anyway , that's not the point , i know how to do that, but i just 
 think it's ugly
btiffin
27-Jun-2007
[1987x2]
parse "ABC" [3 [1 "A" | 1 "B" | 1 "C"]]  ?
nevermind...  :)  AAA returns true
Steeve
27-Jun-2007
[1989x2]
yep
i really think it's not feasible without using counters
Brock
27-Jun-2007
[1991x3]
I think I have it
define a rule;
rule: [1 ["A" | "B" | "C"]]
then parse
parse "cAB" [[rule] [rule] [rule]]
the one indicates that there must be an occurence of one of the three 
characters... supplying the rule then requires three characters to 
be entered.
... but it allows multiple characters... damn
Steeve
27-Jun-2007
[1994]
and it works too with the serie "AAA"
Brock
27-Jun-2007
[1995]
which is bad
Steeve
27-Jun-2007
[1996]
yrp
btiffin
27-Jun-2007
[1997]
parse can change the string as it goes right?  What if you tried 
marking them off with X's as you found them?
Brock
27-Jun-2007
[1998]
I was thinking the same thing, removing each character from a block 
as each rule is processed.
Steeve
27-Jun-2007
[1999]
yeah it's possible but it's like hanging counters, you need to have 
some code to check the result and allowing the parser to continue 
further
Brock
27-Jun-2007
[2000]
so your ideal solution would not use anyting other than parse?