World: r3wp

Grrr, I should make more test before posting...forget last four posts

'regset is now on rebol.org, rest of parser is still in works.

Rebolek...You posted the magic prime 797'th Script.  Well Done!  
:)

Hm thanks, but I was aiming for 800th one... ;)

Yep...I've been waiting...  :)

just a quick idea:

FORALL is implemented as mezzanine function. It calls FORSKIP which 
is mezzanine also. As you can see, it's probably not the fastest 
method. So here's the idea. Cannnot be FORALL rewritten to make it 
faster and is it possible to use PARSE to do this?
So I tried and came up with this simple function:

parall: func [
	'word body
	/local data
][
	data: get :word
	parse data compose/deep [

  some [(to set-word! word) any-type! (to paren! [do bind body :word])]
	]
]

(parall is just a shortcut for parse version of forall).


this is very simple function written in five minutes and not very 
well checked, it needs some additional work (eg. it does not return 
same value as 'forall etc).

So let's do some test (using Ladislav's %timblk.r):
>> n: copy [] repeat i 100 [append n i]

== [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 
25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 4...
>> time-block [forall n [b: n/1 + 1]] 0.05
== 3.623046875E-4
>> time-block [parall n [b: n/1 + 1]] 0.05
== 3.814697265625E-6
>> 3.62e-4 / 3.81e-6
== 95.0131233595801

95x faster? whooo....

and what about bigger block?

>> n: copy [] repeat i 10000 [append n i]

== [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 
25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 4...
>> time-block [forall n [b: n/1 + 1]] 0.05
== 3.540625E-2
>> time-block [parall n [b: n/1 + 1]] 0.05
== 3.7994384765625E-6
>> 3.54e-2 / 3.8e-6
== 9315.78947368421

9000x ? omg...

comments?

the devil is in the details... you need to handle break :)

oh yes, break...didn't use it for years, so almost completely forget 
about it :)

Don't forget continue.

First tip: You don't have to bind the body - it just uses the existing 
binding of the word.

Netx tip: What native does forskip decompose down to? Try using that 
first.

; Try against this, the forskip code with the skip part taken out
forall: func [
    "Evaluates a block for every value in a series."
    [catch throw]

    'word [word!] {Word set to each position in series and changed as 
    a result}
    body [block!] "Block to evaluate each time"
    /local orig result
][
    if not any [
        series? get word
        port? get word

    ] [throw make error! {forall expected word argument to refer to a 
    series or port!}]
    orig: get word
    while [any [not tail? get word (set word orig false)]] [
        set/any 'result do body
        set word next get word
        get/any 'result
    ]
]

BrianH: I know it can be done using while loop in forskip, it was 
more an experiment on using parse as foreach/forall kind of function.

to see how fast it can be

Well, if you can figure out how to make it handle break and continue, 
let us know.

Or for that matter, ports.

If you want to test the speed of parse, replace the any-type! with 
a skip - the forall you are comparing it to doesn't do that test.

What is 'continue?

Sort of like the opposite of break. It breaks one iteration of a 
loop and continues at the top of the next iteration. Many languages 
have it, as does REBOL 3.

notice that, forall was initially done that way, and converted to 
use forskip for simplicity (ie. avoiding having the same code twice 
in the source)

I tried to enclose parse in loop 1 [] and it seems to handle break. 
I guess you'll probably prove me wrong, Brian :)

parall: func [
	'word body
	/loc data
][
	loop 1 [
		data: get :word
		parse data compose/deep [
			some [(to set-word! word) skip (to paren! [do body])]
		]
	]
]

re: continue - this is not r3 ;)


>> n: [1 2 3 4 5] parall n [if n/1 = 4 [break/return "break"] if 
n/1 > 4 [print "bad"]]
== "break"

good :) now... if Ladislav was here he'd probably find at least ten 
wrong things in that function. ;)

you need to add [throw] for example, and return the last return value 
of the body

but... as long as we're not replacing forall... this could be a very 
fast alternative to use in cases when you don't need all the features...

and actually... i'd call parse directly in that case ;)

forall replacement not...an experiment in using parse for maybe not 
obvious purposes - yes :)

You could make something new. like this:
 parall [p: set x integer! set y integer!] [?? p ?? x ?? y]

parall [p: set x integer! set y integer!]  DATA [?? p ?? x ?? y] 
;..

To be able escape from infinite loops use () somewhere in the parse 
rules as for example:

 >> parse "abc" [any [()]]
 *** YOU CAN PRESS [ESC] NOW TO STOP THE LOOP ***

>> parse "abc" [any []]
*** HANGS REBOL ***

Cool!

Oldes: does using () inside a parse loop slow things down? I'm wary 
of () as they do seem to slow the interpreter down in many cases.

Peter it does slow interpreter down, it's better to use this trick 
when designing your rules. If you're happy with them, remove the 
parens to make your rule faster.

What about a new word “all” which would allow  to parse several successives 
conditions whatever their order  ?

parse [a: 1 b: 2] [all ['a: integer! | 'b: integer!]] would be true 
only if it exists successivly [a: integer!] and [b: integer! ] whatever 
their order

need it

Umm...I'm not really a parser but


parse [a: 1 b: 2] [set-word! integer! set-word! integer!] returns 
true.

yeah but what ? if the entry is [b:2 a:  1] instead

hum sorry, it works too

but if the entry is more complex like [a: 1 huge tight  b: 1] and 
that i want parse all this values in any order ?

and in your example , you don't test that we must encounter 'a and 
'b

i give another one example , more simple

i want to parse the serie "ABC" but it could be the serie "BCA" , 
"CBA" , "ACB" ... any combination

so if i could write parse "ABC" [all ["A" | "B" [ "C" ]]  , it should 
be more simple

Again, I'm not a parser, but parse's support of alternates with some 
and any covers these cases no?

no

if you write  parse "AB" [some ["A" |  "B" | "C"]]

parse "ABC" [ 0 1 "A" | 0 1 "B" | 0 1 "C"] ?

you get a true answer, but that not the case

I think I'm clue'in to what you mean.  I'll leave this to the pros... 
 :)

like my use of 0 1 can be opt...  :)

anyway, you're example just don't work ^^

*anyway your example just doesn't work (fewwww)