OpenEuphoria

Re: Why equal(x[n], x[n..n])=0 ?

• Posted by Fernando Bauer <fmbauer at h?tmail.?om> Sep 26, 2007
• 723 views

Hi Ricardo. Thanks for your reply.

Ricardo Forno wrote:
> 
> Hi Fernando.
> Maybe the point is that x[1..5] as well as x[3..3] are slices (two indexes
> separated
> by two periods), while x[5] (a single index) is not a slice.
> The simplest case: assume all elements of x are atoms. Then x[i] is an atom,
> and x[i..j] is a sequence of atoms.
> Assume also that you have this loop (code not tested):
> 
> x = repeat(4, 10)
> 
> for i = 10 to 1 by -1 do
>    foo(x[1..i])
> end for
> 
> The foo() procedure gets sequences with diminishing length, and processes
> them.
> It expects it has to deal with a sequence.
Yes. For functions expecting a sequence, only the current definition of slice
works, and probably this is the cause of that slice definition. But then I ask,
why not define a similar but more generic function which accepts object?

> But, oh surprise!, when you get to i = 1 (if x[n] where the same as x[n..n],
> as you propose), it would get an atom!, and probably fail.
> So, foo() should test if its argument is a sequence or an atom before
> proceeding.
Yes. I think this is a common procedure in all functions that receive objects.

> Don't you think this way things would be more complex?
For now, I don't know. I think that I'd have to have much more experience with
Euphoria to correctly answer this.
I'm learning with all of you with these discussions.

> 
> Best regards.
> 
> PD: Hi Fernando.
> I am from Buenos Aires, Argentina. I recall you are from Porto Alegre, Brazil,
> true?
Yes. And I know a few things about you thanks to the Euforum.

> I understand enough Portuguese. You can contact me at ricardoforno at
> tutopia.com
> in Portuguese if you like.
> Ah! And I plan to take my holydays in the south of Brazil next January, with
> my wife. Maybe we can meet then.
Ok. Maybe.

Best Regards,
   Fernando