From: Robert Klemme <shortcutter@...>
Date: 2008-10-29T19:41:10+09:00
Subject: Re: lambda and multiple arguments: how?!

2008/10/28 nvp <ruby-lists@noopy.org>:
> On Tue, Oct 28, 2008 at 1:36 PM, David A. Black <dblack@rubypal.com> wrote:
>> You could call it as: code.call(pid,*params),
>
> This didn't work for me with:
>
>  lambda {
>   |pid,exit_code,ident|
>  # ...
>  }
>
> What am I missing here?

You need to adjust the parameter list to the arity of the block.
Here's a way how to do it

11:38:30 Temp$ cat lambda.rb
@code = lambda {|a,b,c| printf "R: %p %p %p\n",a,b,c}
def run1(*params)
  ar = @code.arity
  if ar >= 0
    params = params[0...@code.arity]
    params.concat(Array.new(ar - params.size))
  end
  @code[*params]
  @code.call(*params)
end
def run2(*params)
  params.unshift :pid
  ar = @code.arity
  if ar >= 0
    params = params[0...@code.arity]
    params.concat(Array.new(ar - params.size))
  end
  @code[*params]
  @code.call(*params)
end
run1 1
run1 1,2
run1 1,2,3
run2 1
run2 1,2
run2 1,2,3
11:38:32 Temp$ ruby lambda.rb
R: 1 nil nil
R: 1 nil nil
R: 1 2 nil
R: 1 2 nil
R: 1 2 3
R: 1 2 3
R: :pid 1 nil
R: :pid 1 nil
R: :pid 1 2
R: :pid 1 2
R: :pid 1 2
R: :pid 1 2
11:38:33 Temp$

Btw, IMHO you have a slight design inconsistency in your code: the
client registers a lambda with a certain pid so it *knows* the pid
already.  But still you provide it to the block when calling it.  That
seems superfluous because even if the pid is stored in a local
variable the block can access it because it is a closure:

irb(main):001:0> def bl
irb(main):002:1> pid = Time.now.to_i # random
irb(main):003:1> lambda { puts "pid is #{pid}" }
irb(main):004:1> end
=> nil
irb(main):005:0> x = bl
=> #<Proc:0x7ff8e04c@(irb):3>
irb(main):006:0> x[]
pid is 1225276874
=> nil
irb(main):007:0> x[]
pid is 1225276874
=> nil
irb(main):008:0> x[]
pid is 1225276874
=> nil
irb(main):009:0> x[]
pid is 1225276874
=> nil
irb(main):010:0>

Kind regards

robert

-- 
remember.guy do |as, often| as.you_can - without end