Please help me out

Hello All, I have the below; kool: %gin.txt write kool "African" write/append kool "Child" Now, I would want to read the file , and print first "African", and second Child I tried read/lines kool == [ "AfricanChild"] I was looking for something like [ "African" "Child"] Regards, Emeka

If you open the file in a text editor, you'll see that it does contain AfricanChild on one line. All that's needed is a newline in the middle: write kool "African" write/append kool newline write/append kool "Child" Alternatively, 'write has another refinement that could help here: write/lines kool "African" write/lines/append kool "Child" The /lines refinement works similar to read/lines - this would result in the same as the above: write/lines kool [ "African" "Child" ] On Sun, Aug 22, 2010 at 11:46 AM, Emeka <emekamicro-gmail.com> wrote:

> Hello All, > I have the below; > To unsubscribe from the list, just send an email to > lists at rebol.com with unsubscribe as the subject.

Hello All, Is [] a cousin of LISP list () (serving the purpose of procedure call and data structure) Regards, Emeka On Sun, Aug 22, 2010 at 6:09 PM, Izkata <izkata-gmail.com> wrote:

On Sun, Aug 22, 2010 at 11:51 PM, Emeka <emekamicro-gmail.com> wrote:

> Hello All, > > Is [] a cousin of LISP list () (serving the purpose of procedure call and > data structure)

Henrik and Izaka, Thanks so much. More blogs posts won't be harmful. Regards, Emeka On Mon, Aug 23, 2010 at 1:11 PM, Henrik Mikael Kristensen < henrikmk-gmail.com> wrote:

Hi, the answer of Henrik is fine. To answer the "Lisp" part: in Rebol [x y] does not mean: - apply function x to argument y which in the Lisp list (x y) would be the case. [] in Rebol is just data. Regards, Arie 2010/8/22 Emeka <emekamicro-gmail.com>

Arie, But when you apply func to [x y] it seems to me that it starts to behave like Lisp. Am I wrong? Emeka On Mon, Aug 23, 2010 at 1:51 PM, Arie van Wingerden <xapwing-gmail.com>wrote:

No. In Lisp the x would be evaluated as being a function with arg y. In Rebol x is (generally speaking) not a function. Do you have a short example to show your confusion? 2010/8/23 Emeka <emekamicro-gmail.com>

I don't know much about computer science, but I think REBOL's blocks share some things in common with Smalltalk: http://web.cecs.pdx.edu/~harry/musings/SmalltalkOverview.html#Blocks REBOL's cool twist is that a block can hold code or data and treat them both equally/interchangeably, and can be accessed in a uniform manner (a data collection). In LISPy languages, the term 'thunk' might be the appropriate comparison. http://stackoverflow.com/questions/925365/what-is-a-thunk-as-used-in-scheme-or-in-general -- Ed On Mon, Aug 23, 2010 at 10:09 AM, Arie van Wingerden <xapwing-gmail.com>wrote:

Hello All, This where I got my conclusion. I mentioned before that blocks are a container type. An ordered sequence of values. Actually they are an ordered sequence of un-evaluated values. What that means is that when a block itself is evaluated the values it contains are not evaluated. The values a block contains are evaluated when a function is applied to the block http://www.codeconscious.com/rebol/articles/rebol-concepts.html#Firsttrapforbeginners Is the last statement correct? Emeka On Mon, Aug 23, 2010 at 4:09 PM, Ed O'Connor <edoconnor-gmail.com> wrote: I don't know much about computer science, but I think REBOL's blocks share

Indeed that is the way it works. In Lisp a list (when unquoted!) is always evaluated. In Rebol evaluation is not standard, but must be triggered by a calling function (outside the block). And indeed also, when a block is being evaluated functions within the block will be recognized and evaluated also. 2010/8/23 Emeka <emekamicro-gmail.com>

On Mon, 23 Aug 2010 12:10:34 -0400, Emeka <emekamicro-gmail.com> wrote:

> Hello All, > This where I got my conclusion. > is applied to the block > Is the last statement correct?

Why REDUCEd? I am a bit confused now. Emeka On Mon, Aug 23, 2010 at 10:13 PM, Brian Tiffin <btiffin-rogers.com> wrote:

On Mon, Aug 23, 2010 at 11:23 PM, Emeka <emekamicro-gmail.com> wrote:

> Why REDUCEd? =A0I am a bit confused now.

On Monday, 23-August-2010 at 22:23:53 Emeka wrote,

>Why REDUCEd? I am a bit confused now.

>> cat: "feline"

>> dog: "canine"

>> blk: [cat dog 1 + 1]

>> reduce blk

>>> b: [reduce blk 10]

>> foreach value b [print [value "type:" type? value]]

It's a "belief" of mine. Data in blocks are just literal data until a REDUCE. Nifty cool data-type! data, but still just the literal data. Explicit or implicit in other words like DO or paren expressions in COMPOSE, to me REDUCE is one of the magic concepts that lets REBOL look and feel easy while being deep and vast. Now explaining part of the belief. It started dawning on me when I began the trek to figuring out what the heck lit-words were and why REBOL needs them. The rest is just example; ... being unable to explain REBOL in a way that is truthful and not just the look and feel easy bit. a: 42 b: 9 data: [a b] literaldata: ['a 'b] human: ['a 'is a 'and 'b 'is b] ; the type? of these entries is word! probe data [a b] probe reduce data [42 9] ; integer! reduces to the same integer! probe reduce reduce data [42 9] ; the type? of the entries is lit-word!, reduced to word! probe reduce literaldata [a b] probe reduce reduce literaldata [42 9] ; lit-word! mixed with word! probe reduce human [a is 42 and b is 9] ; the word is is not bound to a context that can give a value probe reduce reduce human ** Script Error: is has no value Cheers, Brian On Mon, 23 Aug 2010 17:23:53 -0400, Emeka <emekamicro-gmail.com> wrote:

Brain + Carl + Henrik, Hmm, this is like standing on the shoulders of giants. Thank you all. Emeka On Tue, Aug 24, 2010 at 8:09 AM, Brian <btiffin-rogers.com> wrote:

Dne 23.8.2010 19:26, Arie van Wingerden napsal(a):

> Indeed that is the way it works. > In Lisp a list (when unquoted!) is always evaluated. >> Emeka >>

Ladislav, Your 2 cents worths more than my 50 cents Thanks. Emeka 2010/8/24 Ladislav Mecir <lmecir-volny.cz>

Ouch, my head hurts. Just wanted y'all to know that us lurkers really appreciate these types of discussions.

Hi all, Just in case you missed it in the past, here is some discussion that went between Josh F and Maxim recently and that I found last July 24th. This can be of interest too since it is related - Sorry I couldn't post the link only as I didn't kept the reference, only the contents. ============= = JoshF post : ============= I have a very strange problem with REBOL/View 2.7.6.3.1... If I do this: repeat i 5 [do [print i]] I get five number down the screen as you'd expect. ------------------------------ ----------------- 1 2 3 4 5 However, if I do this: m: [print i] repeat i 5 [do m] Then I get a "** Script Error: i has no value ** Near: print i" error. ----------------------------------------- and Oh, by the way... m: [print i] repeat i 5 m works just fine too. ---------------------------------- As far as I can tell the operations should be identical. Any ideas as to what I'm doing wrong? Thanks! ========================= = Maxim Olivier Adhloch reply : ========================= its a binding issue... you see the i variable is within a block which is evaluated by do, but the binding of the [do m] block is done by repeat. as such, binding does not traverse words. repeat will assign the value of m to your block, but not the values of the words inside of m to i or print. in fact, do is doing this indirectly when its looking at the block. but because i isn't defined outside of repeat... i maps to nothing. now repeat did bind m, but not the values inside of m . this is perhaps the trickiest part of binding. a word and its value are not equal wrt to binding... you will have to reduce the repeat block, so that m is reduced or composed into its value, before repeat can bind 'i . in a very old version of rebol there was "aggressive" binding and it did recurse into words, like you are expecting it, but that caused some impossible to reverse side-effects... so this was removed. in your case, just do: m: [print i] repeat i 5 reduce [ 'do m] ---------------------------------------------------- or m: [print i] repeat i 5 compose/only [ do (m) ] ------------------------------------------------------------------ In fact if I apply the recomendation from Ladislav, using compute instead of eval could have made the real sense easier to grasp from the beginning - at least to me ! Thanks all of you. HTH, Gerard P.S. Now all that is missing for me is some in context use for a real binding study, as could be applied to differents "contexts" - for example when words are not bound to the main R2 dictionary store. This already bogged me down and I have to restart these old projects ... whenI will have more time - just to see if I understand these concepts better now ... after all these long years going up and down myself !

And of course this will do the job: m: [print i] repeat i 5 [do bind m 'i] this will bind the m into the context of i (which is context of repeat) before DOing it. Tue, 24 Aug 2010 15:50:32 +0300 tarihinde G�rard C�t� <gerardcote-gmail.com> ��yle yazm��:

Dne 24.8.2010 14:50, G�rard C�t� napsal(a): ...snip...

> In fact if I apply the recomendation from Ladislav, using compute instead of > eval > could have made the real sense easier to grasp from the beginning - at least > to me !