[REBOL] Re: Scope? Any advice would be appreciated.
From: lmecir:mbox:vol:cz at: 28-Jan-2001 21:07
> Thanks to all that replied!
> Everything somewhat makes sense. Everything is just an "object" that
> maintains itself once declared. And code appears to be self-modifying.
> However, I have a follow-up question. The following routine *does* work:
> lcWorks: func [ /local b ]
> b: 0
> print "The following line s/b 0"
> print b
> b: b + 1
> print "The follwing line s/b 1"
> print b
> It doesn't need to use "copy" to make it work. I guess this issue just
> boils down to assignment techniques.
This issue doesn't have anything in common with the "assignment techniques".
> Why do I have to use different assignment techniques? That is, in the
> above example, a straightforward assignment of 0 to b works as one would
The difference lies in the fact, that the expression b: b + 1 doesn't change
the second value of the LCWORK function's body. In a different wording: the
second value of the LCWORK's body remains unchanged, because it is
Immutable. For more info on Immutable vs. Mutable see e.g.
> However, the same concept applied to a block, b: [ 0 0 0 ] as in
> the earlier example, does not work. Instead I have to make a copy of [ 0
> 0 ] - b: copy [ 0 0 0 ].
> I am not clear as to why I would need to make a copy - why wouldn't the b:
> [ 0 0 0 ] just blow away the current contents, even if the function and
> properties are maintained between calls?
It would, but the problem lies in the fact, that the code looks different
when modified. If you make a copy, any subsequent change will modify only a
copy of the second value contained in the function body.
> Similarly, in one of the previous
> answers from Gabriele, he showed what appears to me to be behavior of self
> modifying code. However, I still don't understand why using "copy" forces
> the code to update, whereas the normal assignment operator does not.
As you might have seen above, quite the opposite is true. If you don't make
a copy, a modification "updates" your code. If you make a copy, no
modification of the original occurs.
> Lastly, since the code appears to be self-modifying, and behavior of
> assignments change depdending upon whether it is the first call or a
> subsequent one, I am curious as to what advantages this has vs the
> confusion that it will cause people coming from other programming
> Thanks again,
> Victor Mascari
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