Mailing List Archive: 49091 messages

## [REBOL] Re: reduce/deep

### From: joel:neely:fedex at: 21-Oct-2003 15:39

```
Hi, Robert,

Robert M. Münch wrote:
> ;-) But this doesn't seem to work for nested blocks:
>

Sure it does!

>>>d: reduce [a b]
>>
> == [[a 1 b 2] [a 1 b 2]]
>
>>>e: reduce [a b]
>>
> == [[a 1 b 2] [a 1 b 2]]
>
>>>same? d e
>>
> == false
>

Consider a simpler, analogous set of evaluations:

>> a: b: "xyz"        == "xyz"
>> same? a b          == true
>> c: reduce [a b]    == ["xyz" "xyz"]
>> d: reduce [a b]    == ["xyz" "xyz"]
>> same? c/1 d/1      == true
>> same? c/2 d/2      == true
>> same? c d          == false
>> equal? c d         == true

or even *more* simpler (pardon the grammar!  ;-)

>> p: "12"          == "12"
>> q: "12"          == "12"
>> same? p/1 q/1    == true
>> same? p/2 q/2    == true
>> same? p q        == false
>> equal? p q       == true

These all illustrate the difference between SAME? and EQUAL? in that
it is entirely possible to have two different series values whose
corresponding elements are the same.  In my first example above,
as in your original post, two different REDUCE expressions over two
different blocks will not produce THE SAME block, even if the content
of those two blocks are the same.  However the blocks are equal.

My second example breaks it down even further.  Two different
appearances in the input of

"12"

correspond to two different strings, even though the corresponding
characters in those strings are equal (and same, being immutable).

Likewise, two different appearances of

[a b]

result in the creation of two different blocks, even though the
corresponding words in those blocks are the same.  Therefore, when
we reduce those distinct but equal blocks, we get distinct but equal
resulting blocks.

-jn-

--
----------------------------------------------------------------------
Joel Neely            joelDOTneelyATfedexDOTcom           901-263-4446

Enron Accountingg in a Nutshell: 1c=\$0.01=(\$0.10)**2=(10c)**2=100c=\$1
```