World: r3wp

i think that is hard to do perfectly. for example "\"", how to parse 
that?

in actionscript it's doSomeFunction("a" add b, 1)

doSomeFunction("a" add b # 1)
doSomeFunction("a" add b . 1)
?

I know it woud be possible to improve my dialect, but I'm not going 
to do it in near future, I'm already used to write it this way:)

I think it's quite logical to close it in parenthesis than using 
chars like #

But chars like # are easier to search/replace. But i dont defend 
my solution, yours looks ok.
Do you write your dialect-code in rebol-blocks or in a string?

And for the Rebolek: I use in RSWF this:

slash: to-lit-word first [/]
	dslash: to-lit-word "//"
	rShift: to-lit-word ">>"
	UrShift: to-lit-word ">>>"
	_bigger: to-lit-word ">"
	_less: to-lit-word "<"
	_noteql: to-lit-word "<>"
	_lesseql: to-lit-word "<="
	_biggeql: to-lit-word ">="

and then:

'* (term-op: [ins-act #{0C}])
				| slash  (term-op: [ins-act #{0D}])	;Divide
				| dslash (term-op: [ins-act #{3F}])	;Modulo
				| rShift (term-op: [ins-act #{64}])	;RShift
				| UrShift (term-op: [ins-act #{65}]);UnsignedRShift
				| lShift (term-op: [ins-act #{63}])	;lShift
				| '| (term-op: [ins-act #{61}])	;bitwise OR
				| ['|| | 'or] (term-op: [ins-act #{11}]);OR
				| ['& | 'band] (term-op: [ins-act #{60}])	;bitwise AND
				| 'and (term-op: [ins-act #{10}])	;AND

Thanks Oldes, that should help me

Volker: blocks

it's much more easier as you can use Rebol's datatypes

i thought about something like "js-load string" which would do the 
comma-conversion. But if your dialect works, no need for extra work.

maybe fixing REBOL should be better? If "." can work without problem, 
why not "," ?

interesting question why it is left out. but you would have other 
problems too, although not that often. like "\"". in rebol that is 
"\" " . in javascript "^""

Volker this is some special case, you can replace that before parsing 
string, but you cannot replace all commas in action script, some 
of them may be parts of string

Does anybody have an script to parse BNFs?

BHF's specification..

I think Bretty Handley did. Check codeconscious.

I think this should print 5 and return 'true. But it does not.

>> parse [1 2 3 4 5][any number! set val number! (print val)]
== false

Help please? :)

>> parse [1 2 3 4 5][any [set val number!] (if value? 'val [print 
val])]
5
== true

What you did was consume the five numbers, then try to set val to 
a sixth one.

What I want is work with first four numbers and not with the last 
one

Is it always 5 numbers ?

No :(

>> parse [1 2 3 4 5][4 [set val number! (print val)] number!]
1
2
3
4
== true

Actually, the real elements may be very different. I just simplified 
the example.

Ok, so you don't know how many numbers you have until you fail to 
find another one.

I want to process all elements excluding the last one.

And do you want to avoid putting them into a block first ?

They are in block already

Actually, you can do it like this:

>> parse [1 2 3 4 5][any [set val number! pos: number! (print val) 
:pos] number!]
1
2
3
4
== true

Interesting, that should probably work, thanks!

Welcome.

Actually, the last   number!   can probably become   opt number!

for the case when there are no numbers at all.

there is at least one. I f theres at least one, don't do any action. 
If there are two do one action and so on.

Ok, that should be ok then.

Man, I wish you could do:

	parse [1][integer! -1 skip]

and arrive back at the head of the input.

infinitive loop?

(infinite)

Another possible way:

>> parse [1 2 3 4 5] [any [set val number! pos: (if not tail? pos 
[print val])]]

Anton, you can do:
>> parse [9] [integer! to 1]
and arrive back at the beginning.

Oh yes! forgot about that. :) Great !

Can you explain this curious results ?

REBOL/View 1.3.2.3.1 5-Dec-2005 Core 2.6.3

>> parse [1 2 3 4][any [number! set val number!] (print val)]
4
== true
>> parse [1 2 3 4 5 ][any [number! set val number!] (print val)]
4
== false
>> parse [1 2 3 4 5 6][any [number! set val number!] (print val)]
6
== true

>> parse [1 2 3 4 5 6 7][any [number! set val number!] (print val)]
6
== false

>> parse [1 2 3 4 5 6 7 8][any [number! set val number!] (print val)]
8
== true

note the results with odd numbers of items!

Forget my  question. I see that the block tries to consume two items.(

:)

'parse' is the path to great explorations and inventions - and also 
to great confusion and maybe despair. ;-)


No really, it can be a bit confusing at times, but I guess, it can't 
be done otherwise to have such great functionality. There's no short 
cut with 'parse'. Learning by doing is the way to go. And it's a 
brilliant tool!