lambda callee

[Marcin Mielżyński]

Hi,

I wonder if something like this can be done without giving lambda a name:

(l=lambda{|a|a.zero? ? 1: a*l[a-1]})[10]

something similar like javascript arguments.callee comes to my mind..

lopex

 

[Pierre Barbier de Reuille]

Jeffrey Schwab a écrit :

Hi,

I wonder if something like this can be done without giving lambda a name:

(l=lambda{|a|a.zero? ? 1: a*l[a-1]})[10]

something similar like javascript arguments.callee comes to my mind..

lopex

Skip the recursion. Instead of storing each successive value on the
call stack, store it using inject's accumulator.

(lambda {|a| (1..a).inject {|p,n| p * n } })[10]

Btw, for the record, the variable name l is horrid to begin with, and
interspersing it with the number 1 in a one-liner might make you go
cross-eyed.

Another solution (because the inject method is not *always* possible to
use) is the Y operator, which is very simply defined in Ruby :

def y(&f)
lambda { |*args| f.call f,*args }
end

And used with :

y { |f,a| a.zero? ? 1 : a*f[f,a-1]}[10]

As you can see, the block of the Y operator will get itself as first
argument.

Pierre