OpenEuphoria

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

• Posted by Andy Drummond <andy at kestrelte?e.co?> Sep 24, 2007
• 711 views

Fernando Bauer wrote:
> 
> Hi All,
> 
> I was debugging a function when I noticed that a slice with equal indexes is
> different from an access with one index. x[n] is different from x[n..n] where
> n is a valid index.
> According to the manual, a slice always result in a sequence (also when x[n]
> is an atom). But, in particular, when we have something like x[n..n], the
> result
> is x[n] with the its depth incremented by 1. In other words, the structure of
> the element depends on the access form (subscription or slicing). For me, this
> is a surprising fact. 
> This kind of implementation affects some algorithms, because we have to test
> when the indexes are equal in order to not use slice.
> So, why equal(x[n], x[n..n])=0 ?
> 
> Regards,
>    Fernando

At a rough guess I'd say x[n] is an atom, the n'th item in the sequence,
(assuming a sequence of atoms), whereas x[n..n] is a one-element sequence.
So you're comparing an atom and a sequence. Essentially the first is an item
from the sequence, the second is a sequence containing the item.

Andy