> -----Original Message-----
> From: Damian Conway [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, May 31, 2005 11:14 PM
> To: perl6-language@perl.org
> Subject: Re: reduce metaoperator on an empty list
>
> 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'
Joe Gottman
