From: Joel VanderWerf <joelvanderwerf@...>
Date: 2010-09-16T05:45:13+09:00
Subject: Re: access super from yield block?

On 09/15/2010 01:12 PM, 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.
>
> # simplified...
> class A<  B
> ...
>
> def parse
>     do_parse {|a| super a}
> end
>
> def parse!
>     do_parse {|a| super a}
> def
>
> def do_parse
>     argv = gets.strip.to_a
>     return ['-x'] if argv.index '-x'
>     return ['-h'] if argv.empty?
>     yield argv
> end
> protected :do_parse
>
> ...
> end

That should work. Can you isolate the problem in a small piece of code 
you can post here?

Here's a working example:

class A
   def foo a
     "A#foo: #{a}"
   end
end

class B < A
   def do_something
     yield 3
   end

   def foo
     s = do_something {|a| super a}
     "B#foo: " + s
   end
end

p B.new.foo # ==> "B#foo: A#foo: 3"