From: 7stud -- <bbxx789_05ss@...>
Date: 2007-10-28T05:43:56+09:00
Subject: Re: .each do |foo, bar| what does bar do?

Brian Adkins wrote:
> On Oct 27, 7:05 am, "David A. Black" <dbl...@rubypal.com> wrote:
>>        i += 1
>>      end
>>      self
>>    end
>> end
> 
> Or this:
> 
> class Hash
>   def each
>     each_key {|key| yield key, self[key] }
>   end
> end
> 


That suffers the same problem as David Black's example.


> I know, you meant w/o recourse to each* :)

My tests show that each_keys() does not call Hash#each(), so your 
example seems to use fair means to me:

class Hash
  alias :orig_each :each

  def each(&block)
    orig_each(&block)
    puts "orig each called"
  end

  def my_method
    each_key {|key| yield key, self[key] }
  end
end

h = {"a"=>1, "b"=>2}

#call original each() method for a hash:
h.each do |key, val|
  print key, " ", val
  puts
end

puts

#call a method that uses each_key():
h.my_method do |key, val|
  print key, " ", val
  puts
end


--output:--
a 1
b 2
orig each called

a 1
b 2

Note that in the last output Hash#each() wasn't called.

That example has raised a question of my own.  Instead of having to 
write:

 def each(&block)
    orig_each(&block)

why can't I relay the block to orig_each() without the second '&', like 
this

 def each(&block)
    orig_each(block)

According to pickaxe2, p56, the '&' method converts the specified block 
to a Proc object and assigns it to the parameter variable 'block'.  Why 
is the second call to '&' required?
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