red/help

For help with your code, suggestions and ideas, and learning Red.

GiuseppeChillemi

@GiuseppeChillemi

Hi, again on paths:

Is it possible to store a path in a word and have it evalued later ?

nedzadarek

@nedzadarek

@GiuseppeChillemi w: 'foo/bar/baz

GiuseppeChillemi

@GiuseppeChillemi

example: path a/b/(c)/d/e

nedzadarek

@nedzadarek

foo: [bar [baz 42]]
w: 'foo/bar/baz
get w
; == 42
reduce w
; == 42

w2: 'foo/('bar)/baz
; == foo/('bar)/baz
get w2
; == 42

GiuseppeChillemi

@GiuseppeChillemi

could I "select" ?

nedzadarek

@nedzadarek

select?

w3: 'foo/('bar) 
select get w3 'baz

GiuseppeChillemi

@GiuseppeChillemi

Headache !

nedzadarek

@nedzadarek

What's wrong?

(sorry, it's midnight - I don't have fresh mind)

GiuseppeChillemi

@GiuseppeChillemi

Midnight here too...

I suppose this syntax is not comatible with R2

nedzadarek

@nedzadarek

Let me try

GiuseppeChillemi

@GiuseppeChillemi

I suppose

select w3 'baz

Is not possible

w3: to-path w3
select w3 'baz

Is not possible too

nedzadarek

@nedzadarek

@GiuseppeChillemi get doesn't allow path! and reduce doesn't work... I'll check alternatives.

^^ yes, not possible

@GiuseppeChillemi hmm.. this work:

>> ? :w
FOO/BAR/BAZ is an integer of value: 42

GiuseppeChillemi

@GiuseppeChillemi

It is possible to use the

w: 'FOO/('REDUCES-TO-BAR)/BAZ
? :w

@nedzadarek

w: 'FOO/(:REDUCES-TO-BAR)/BAZ

is the answer

(REBOL 2)

nedzadarek

@nedzadarek

@GiuseppeChillemi I found that reduce does not reduce path (I am not sure why) but it is happy with a block:

type? first [w]
; == word!
type? first reduce [w]
; == path!
 first reduce [w]
; == foo/bar/baz
first reduce reduce [w]
; == 42

GiuseppeChillemi

@GiuseppeChillemi

I have enough information...

But still have to understand why this works:

foo/('bar)/baz

nedzadarek

@nedzadarek

@GiuseppeChillemi do you mean paren! (()) and 'bar inside it?

GiuseppeChillemi

@GiuseppeChillemi

Yes

hexpected paren! () and :bar

*expected

nedzadarek

@nedzadarek

As parens in the paths are evaluated you will end up with the value that bar points to. What you want is word! (or any-word! - more later if you need additional informations*).

* it is all the same for red if you want select something:

>> arr: [a 42]
== [a 42]
>> select arr to-word "a"
== 42
>> select arr probe to-word "a"
a
== 42
>> select arr probe to-get-word "a"
:a
== 42
>> select arr probe to-set-word "a"
a:
== 42
>> select arr probe to-lit-word "a"
'a
== 42

GiuseppeChillemi

@GiuseppeChillemi

@nedzadarek
To me 'something mean "literal word", so no-translation

nedzadarek

@nedzadarek

@GiuseppeChillemi it is not that simple. At "top" level it says "do not evaluate this word", a: 'b. You will end up with a word!: ? a. In a block words/pathes are not evaluated so you will end up with lit-word! (type? first ['a]).

GiuseppeChillemi

@GiuseppeChillemi

Totally obscure at the moment

Needs more information

Needs input ! Beep !

nedzadarek

@nedzadarek

@GiuseppeChillemi What do you do not understand?

GiuseppeChillemi

@GiuseppeChillemi

@nedzadarek

@GiuseppeChillemi it is not that simple. At "top" level it says "do not evaluate this word", a: 'b. You will end up with a word!: ? a. In a block words/pathes are not evaluated so you will end up with lit-word! (type? first ['a]).

The whole explanation

Vladimir Vasilyev

@9214

@GiuseppeChillemi can you reformulate your question once again?

nedzadarek

@nedzadarek

@GiuseppeChillemi
When you set words to some values, <word>: <value> (e.g. the-answer: 42). <value> will be evaluated.

integer! (e.g. 42) will give you 42 (the value looks the same).
float! (e.g. 3.14) will give you 3.14
string!, pair! (2x14) and some more - like above

There are some "special" cases:

path! (e.g. foo/4) will be evaluated. For example you have foo: [1 2 3 4], when you set value to some word, val: foo/4, path foo/4 will be evaluated. foo/4 will be changed, in this case, to the 4th value of the foo (4).
lit-path! (e.g. 'foo/4) will be evaluated to path! foo/4. As you know already, you can get or reduce it. w: foo/4 and get w will give you 4 (4th element of the foo).
get-path! will get you value that it points to. It is like path!. However with path! you cannot get function!s for example (funs: #() funs/not-fibo: func [a] [ a - 10], funs/not-fibowill try to call not-fibo in the funs map, it will raise an error because you haven't provided an argument). You can get such value by using lit-path!: my-fun: :funs/not-fibo. After this you can call it like any other function: my-fun 12

block! every value within a block will be "the same", it won't use "special" rules that govern path!, lit-path! and get-path!. For example you have a block my-block: [42 3.14 "string" word :get-word 'lit-word set-word: [another block with a word]], you can see that *-path are not evaluated:
Helper:

foreach element my-block [
  print "Value: "
  probe element

  print "Type: "
  print mold type? element

  print "####"
  ]

Output:

Value: 
42
Type: 
integer!
####
Value: 
3.14
Type: 
float!
####
Value: 
"string"
Type: 
string!
####
Value: 
word
Type: 
word!
####
Value: 
:get-word
Type: 
get-word!
####
Value: 
'lit-word
Type: 
lit-word!
####
Value: 
set-word:
Type: 
set-word!
####
Value: 
[another block with a word]
Type: 
block!
####

And last value [another block with a word] will not evaluate words inside (you have no errors because there is no another and a word defined).

btw. tested on Red for Windows version 0.6.3 built 26-Mar-2018/1:14:22+02:00

Vladimir Vasilyev

@9214

@GiuseppeChillemi, @meijeru extensively documented evaluation rules here, if that's what you're interested in.

lepinekong

@lepinekong_twitter

Would it be easy to create a mathematical animation like this in red :)
https://i.imgur.com/IOJSlj0.jpg
https://www.youtube.com/watch?v=-X49VQgi86E

GiuseppeChillemi

@GiuseppeChillemi

@9214 , long day off work, I'll look at your exolaniotion this night or Tomorrow Morning

Giuseppe Chillemi

@GiuseppeChillemi_gitlab

@nedzadarek
Maybe yesterday we were after midnight but my question was different:
This is your code:

foo: [bar [baz 42]]
w: 'foo/bar/baz
get w
; == 42
reduce w
; == 42

I need something like:

z: bar
foo: [bar [baz 42]]
w: 'foo/(z)/baz
get w
; == 42
reduce w
; == 42

(I am just closing my day at work, I'll read it later)

Vladimir Vasilyev

@9214

@GiuseppeChillemi_gitlab perhaps you meant z: 'bar. Aside from that your example already works as you want.

Giuseppe Chillemi

@GiuseppeChillemi_gitlab

Yes ! z: 'bar

So if I write: z: 25, the whole path will be reduced to 'foo/25/'bar

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