On Wed, May 27, 2009 at 05:42:58PM -0500, John M. Dlugosz wrote: > Mark J. Reed markjreed-at-gmail.com |Perl 6| wrote: >> On Wed, May 27, 2009 at 6:05 PM, John M. Dlugosz >> <2nb81l...@sneakemail.com> wrote: >> >>> And APL calls it "|¨" (two little dots high up) >>> >> >> Mr. MacDonald just said upthread that the APL reduce metaoperator was >> spelled "/". As in: >> >> +/1 2 3 >> 6 >> >> So how does |¨ differ? >> >> >> > > Sorry, the two dots is APL's equivilent of the hyper operators, not the > reduction operators. Easy to get those mixed up! > > For example, |1 2 3 ⍴¨ 10| would natively be written in Perl 6 as |10 > »xx« (1,2,3)|. > > --John
Yes. The full expression in raw APL for n! is: */<i>n (where <i> is the Greek letter iota - <iota>n is Perl's 1..$n). Like many things in APL, having a 3-character combination of raw operators to provide a function makes creating a special operator unnecessary. (Although, if I recall correctly, there were also raw operators for combinatorial pick and choose, and factorial(n) is the same as choose(n,n) [or pick(n,n) whichever is the one the considers the order of the returned list to be significant] and factorial was actually returned if there was only one operand provided.)
