Hi Silvia,
Your expression:
set a ""
is evaluated like this, (using parens to indicate current focus):
(set) a "" ; I am a word SET, my value is a function, so let's do
it.
; This function expects two arguments, so let's get
them first:
set (a) "" ; I am a word A, my value is a word B, so let's return
B.
set a ("") ; I am a string, my value is myself, so let's return
that.
; Now SET has all the arguments it needs, so it is done.
So the above is as if you typed:
set 'b ""
which is evaluated like this:
(set) 'b ""
set ('b) "" ; lit-word 'B evaluates to the word B.
set 'b ("")
So the overall problem, I think, is that you are expecting the SET
function to take its WORD argument literally. What actually happens
is that the rebol interpreter reduces A before SET gets to even
look at it.
If you did:
print a
you might think it is PRINT which reduces A to its value, but, in
fact, A is reduced by the interpreter *before* it is passed to PRINT.
The way I think about literal words, words and values is like this.
lit-word -> word -> value
where "->" means "reduces to", and automatic evaluation usually
means moving one step to the right is very easy.
So, in your case:
'b (lit-word) -> b (word) -> "" (value)
and, you have also set A's value to the word B:
a (word) -> b (word) -> "" (value)
Anton.
> Hi everybody!
>
> I know it can sound a bit silly, but I'm going crazy trying to understand
> how exactly work litteral words.
> Now if I do this:
>
> >> set a ""
>
> == ""
> Please let me understand, I'm totally stuck!
>
> Thank you, Silvia
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