Chaim Frenkel <[EMAIL PROTECTED]> writes:
> >>>>> "AS" == Ariel Scolnicov <[EMAIL PROTECTED]> writes:
>
> >> If the original list has no elements, C<reduce> immediately returns C<undef>.
>
> AS> I like everything except this part. Reducing an empty list should be
> AS> an error.
>
> AS> Returning undef (or anything else, really) breaks the algebraic
> AS> equivalence that
>
> AS> f((reduce \&f, LIST), $x) eq (reduce \&F, LIST, $x)
>
> I don't see it.
>
> 1 == f((reduce +, undef), 1) == reduce +, undef, 1
You're confusing "f()" and "+" here, and (on the LHS) "undef" with
"()". What you're claiming is that
f(undef, 1) eq 1
for any function f(). While this is true (in Perl) for C<__+__>, it's
not even true for C<__*__>. And of course there's nothing you can do
about user-specified functions f().
> I really would like to be able to pass around an empty list and get
> a _reasonable_ answer. Having to check the list for emptyness before
> passing it in just seems dirty. One doesn't have to do this for anyother
> looping operator in the language.
>
> for (@empty) {}
> grep {} @empty
> map {} @empty
>
> etc.
Think of the first element of the list as different from the rest --
it is the initial value to reduce from (for + and *, you'll usually
pick an appropriate identity element). By writing
@sum = reduce __+__ 0, @numbers
you deal elegantly with both cases.
NOTE: I find this trick very elegant. I wish it were my trick,
instead of Damian's...
[...]
--
Ariel Scolnicov |"GCAAGAATTGAACTGTAG" | [EMAIL PROTECTED]
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