From: Brian Candler <b.candler@...>
Date: 2010-09-16T06:00:23+09:00
Subject: Re: access super from yield block?

Gene Angelo wrote:
>> What exactly are trying to achieve?
> 
> I have class A that derives from class B
> 
> A#parse and A#parse! override members of the same name in B
> 
> A#parse and A#parse! both call A#do_parse to retrieve arguments from the 
> command line in a loop. A yield in A#do_parse returns the arguments 
> (argv). I want to call 'super argv' from both A#parse and A#parse! but 
> super is not set to A#super.

There is no such method as A#super. super means "call the same named 
method in the superclass".

I tried to make your example into a standalone one, but I don't know if 
this is a legitimate demonstration of what you're trying to do:

----- 8< -----------------------------
class B
  def parse(*args)
    puts "parse in B(#{args.inspect})"
  end
end

class A < B
  def parse
    do_parse { |a|
      puts "Before super"
      super a
      puts "After super"
    }
  end

  def do_parse
    yield [1,2,3]
  end
end

A.new.parse
----- 8< -----------------------------

This prints:

Before super
parse in B([[1, 2, 3]])
After super

which is what I expect. I'm using ruby 1.8.7 (2010-01-10 patchlevel 249) 
[x86_64-linux]

What do you want it to do instead?
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