On Wed, 1 Jun 2005 13.19, Joe Gottman wrote:
> > Juerd asked:
> > >> 2+ args: interpolate specified operator
> > >> 1 arg: return that arg
> > >> 0 args: fail (i.e. thrown or unthrown exception depending on use
> > fatal)
> >
> > > Following this logic, does join(" ", @foo) with [EMAIL PROTECTED] being 0
> > > fail too?
> > No. It returns empty string. You could think of C<join> as being
> > implemented:
> > sub join (Str $sep, [EMAIL PROTECTED]) { reduce { $^a ~ $sep ~ $^b }
> > "", @list
> > }
> >
> > Just as C<sum> is probably implemented:
> >
> > sub sum ([EMAIL PROTECTED]) { [+] 0, @list }
> If this were the case, then
> join '~', 'a', 'b', 'c'
> would equal '~a~b~c' instead of 'a~b~c'
Another broken symmetry. Ah, Perl 5's full of them, and I see that Perl 6
will be no exception. Excuse me while I find my cynic hat . . ah, here it
is.
I'm still in the camp of those wanting each operator to know its own identity
value (perhaps in terms of a trait). The identity of multiplication (say) is
always 1, after all, and it doesn't change depending on when you do
multiplication in your own code. In mathematical terms, it's a property of
the operator, not of how it's used.
You are going to see empty lists more often than you think in expressions like
$product = [*] @array;
and having to write that as
$product = [*] 1, @array;
just to protect against a common case doesn't exactly flaunt Perl's DWIMmery
to me. I *have* to write 1 there, or otherwise the reduce meta-operator
isn't calculating the product when there are items in @array. This hardly
makes sense from a Huffman perspective. Someone please convince me
otherwise.
Multiplication's an easy example, in any case, and any programmer will know
the identity is 1. But look at the confusion here on this list among people
over the identities for <, >, ** and other beasts. I've already forgotten
some of them.
(On another note: I had a thought the other day: [op] produces one thing from
a list, and »op« produces a list from a list. When I squint with hindsight,
it makes more sense to me for »op« to be the one that converges a list to a
single point, and [op] to do listy things to lists. I'm not seriously
suggesting a switch - there's probably parsing issues with the thought - I'm
just sayin', is all.)
--
Debbie Pickett
http://www.csse.monash.edu.au/~debbiep
[EMAIL PROTECTED]
