World: r3wp

sorry, cat on keyboard :)

unfortunately the cat did not type a useful parse rule.

huhu

I thought the zen answer was imminent...

steve - not sure this should be a new dialect word, since it is a 
very specific case. basically you want all permutations of a set 
of rules.

the simplest way may be to just generate all the permutations.

that is, parse string [a b c | a c b | c a b | c b a | b c a | b 
a c]

i give another one example: imagine you have a dialect which describes 
a rectangle with coordinates length and color.

You could write [rectangle 10x20 300 red] but [rectangle 30 red 10x20] 
should be correct too (and other combinations).
So a parser should handle this with not too much complication.

that can also be written as: parse string [a [b c | c b] | b [c a 
| a c] | c [b a | a b]]  (faster)

steve, we do that all the time, and we just take the last value when 
it is put more than once.

Steeve: In that situation, you have a little discretion to ignore 
excess attributes, as VID does.

but what if i have 10 rules to complete ?

that is, two pairs is not an error, the parser just ignores one. 
if vid errored out it would be worse.

you make a function that taken a list of rules generates all the 
permutations.

ok, it was a bas example

*bad

yes Gabriele, it's what i do currently

Not bad, we're just forgiving here : )

but it's uggly

i don't see any better approach, because you are parsing for the 
permutations.

and it's become huge if i have lot of rules to combinate

if you use the above approach (second example i wrote) it is reasonably 
fast.

otherwise, you need self-modifying rules, or use flags.

yes we talked about other solutions

but i have just posted here because i thought it will be a good improvment 
to have a new word in parse dialect

the point is, this has just been asked once in 7 years, so not sure 
how common ;)

but my problem is not so common, it seems...

I haven't asked, but I have come across this too.

The workarounds do seem a little unwieldy, but I'm not sure what 
the complexity in the proposed 'all keyword would be...

internally it would probably have to check flags or do the permutations 
on the fly.

this is possible, but my guess, that Steeve would call it ugly too:


>> once-only: func [rule] [use [rule'] copy/deep [rule': :rule [rule' 
(rule': [end skip])]]
]

>> rule: [(a': once-only a b': once-only b c': once-only c) 3 [a' 
| b' | c']]

== [(a': once-only a b': once-only b c': once-only c) 3 [a' | b' 
| c']]

Ladislav, that's not ugly, it's beautiful! :-)

I have a question though. Do you need the skip at the end of once-only? 
Wouldn't it work just with
rule': [end]

To do block parsing, I suggest:

>> once-only: func [:rule] [use [rule'] copy/deep [rule': :rule [rule' 
(rule': [end])]]]

>> rule: [(a': once-only 'a b': once-only 'b c': once-only 'c) 3 
[a' | b' | c']]
>> parse [c b a] rule
== true

hmm, the block parsing give wrong result, if there are less than 
3 values to parse.

John: [end skip] is a rule that surely fails, while [end] may succeed 
(at end)

I see. That's also why my block parsing give false result with less 
input. So for block-parsing, it should be:

>> once-only: func [:rule] [use [rule'] copy/deep [rule': :rule [rule' 
(rule': [end skip])]]]

why, my rule should be usable as-is even for block parsing?

(I don't like fetched arguments)

>> once-only: func [rule] [use [rule'] copy/deep [rule': :rule [rule' 
(rule': [end skip])]]]

>> rule: [(a': once-only 'a b': once-only 'b c': once-only 'c) 3 
[a' | b' | c']]
>> parse [a b c] rule
** Script Error: Invalid argument: a

in that case you can use either:

    a': once-only first ['a]

or

    a': once-only ['a]

Yup, works. Nice! :-)

Once (again)

Rebol is amazing, i think i found a simple and elegant dialect

Currently, it's not allowing recursive once usage, but it's obvious 
to do with a stack.


take: func [r] [n: 0 once/1: (length? r) + 1 / 3 once/2/2: r replace/all 
r '.. '.]
.: [(n: n + 1)]
..: [(n: n + 1) end skip]
once: [0 [(if n > 0 [poke once/2/2 n - 1 * 3 + 1  '..] n: 0) []]]


rule: [. "a" | . "b" | . "c"]
parse "CBA" [ (take rule) once]
== true
parse  "BAC" [ (take rule) once]
== true
parse  "CBA" [ (take rule) once]
== true
parse  "BBC" [ (take rule) once]
== true
parse  "CA" [ (take rule) once]
== false
parse  "CABA"[ (take rule) once]
== false
rule2: [. "a" | . "a" | . "b" | . "c"]
parse "CABA"[ (take rule2) once]
== true

(correction) 
>>parse "BBC" [(take rule) once]
== false

gnarly

:)

gnarly ?

ok,    http://www.urbandictionary.com/define.php?term=gnarly

data: "cab"
>> c: charset "abc" parse data [3 c]
== true

?

>>>parse "ccc" [3 c]
==true