What does "loop" return?

== [3 3]

R is a block of value: [3 3]

== 3

R is an integer of value: 3 In first case "loop" returns block (seems like ...?) while in second it returns integer (seems like last evaluated value). So there are some questions: - is that desirable ? - what is the sense of such behaviour ? - why only second case is described in REBOL/Core doc ? (and for REBOL 3 too) - if I am wrong and in first case "loop" returns value too then how this value (block [3 3]) is calculated ? Thanks in advance. I can't go ahead without understanding base things. Valeri

== [3 3]

== [3 3]

== [3]

== [3 3] In general, think about what REBOL returns when a block is evaluated. What are you doing when you use LOOP, or FUNC? What does a function return (assuming you don't use EXIT or RETURN in the body)? REBOL is great fun, but it is also very different from other languages. Don't be discouraged by things that seem odd, and feel free to post any questions, even if they seem simple. -- Gregg

For example: series of all evaluated values ;-) So loop 2 [append [] 3] == [[3] [3]] and loop 2 [3] == [3 3]

Why "append append [] 3 3" not "append [] 3 append [] 3"? I do not understand this point. Valeri=

-- Graham Chiu http://www.synapsedirect.com Synapse - the use from anywhere EMR.

Yes, I see what is happening. I read code: loop 2 [append [] 3] as "do 2 times block of code [append [] 3]". Apparently this is not right. Could you please suggest some rule for reading? Valeri=

The first time 3 is appended to the empty block. The second time 3 is appended to the now non empty block which contains 3 you end up with [ 3 3 ] If you wish to create a new empty block each time, you would do this instead loop 2 [ append copy [] 3 ] which would be the same as append [ ] 3 append [ ] 3 -- Graham Chiu http://www.synapsedirect.com Synapse - the use from anywhere EMR.

Thanks for this example. Sorry, now I understood nothing. Why loop 2 [append [] 3] is equivalent to append append [] 3 3 that is some sort of "nested" expressions - result of first 'append is INSERTED as parameter for second 'append, while loop 2 [ append copy [] 3 ] is equivalent to append [ ] 3 append [ ] 3 that is some sort of "sequence" expressions - first 'append is FOLLOWED by second 'append? Why does not parameter substitution done? Valeri

20

You are entering a deep issue in REBOL, namely that certain values are aggressively reused and that REBOL works by copying as little data as possible by default to let it be very efficient. That might be a bit hard to understand at this point. Let's look at the loop: loop 2 [append [] 3] And strip the stuff in front of the block: [append [] 3] When you specify the loop like this, REBOL immediately recognizes the [] you specify as being literally the same memory position that APPEND should work on, each time it runs through the loop. This is very important to understand. In a sense the loop becomes self modifying. This is a neat trick that lets you avoid referencing the block from a word, but it can also be a nasty trap, if you don't realize this. So when you try this one: append [] 3 append [] 3 You have two different blocks at two different memory locations. When you do a COPY, the block you work on is never the same, but always a copy. This trick works everywhere, in blocks, loops, functions, etc. But be careful when you use it. REBOL has a function for testing the SAMEness of series/blocks, i.e. whether two series you are testing on, is the same one at the same memory location. It's called SAME?. -- Regards, Henrik Mikael Kristensen

Thank you very much, Henrik! Valeri PS Do you know more tricks that anyone must know BEFORE start real coding? :-)