From: Lou Zell <lzell11@...>
Date: 2008-08-01T05:51:04+09:00
Subject: Re: super with block

Lou Zell wrote:
> Mikael, I can deal with that, I'll swap out the $'s with block params. 
> Thanks for the workaround!
> 
> Lou

Actually, this did not work the way I expected.  It is fine for a case 
like this:
str = "match1"
gsub!(/(match1)/) {|m| m}

But not this:
str = "match1match2"
gsub!(/(match1)(match2)/) {|m1,m2| m1 + m2}

Where I am really looking for:
str = "match1match2"
gsub!(/(match1)(match2)/) { $1 + $2}


So let me ask this instead:  How do I pass a block from one method to 
another without invoking the block?

For instance, let's say I want to create a method nsub! that behaves 
EXACTLY like gsub!

class String
  def nsub!(*args)
    gsub!(args[0]) {yield}
  end
end

This will not work, say I do this:

s = String.new("whatever")
s.nsub!(/(what)/) {$1}

The yield in the body of nsub! invokes the block {$1}, but in this 
context $1 is nil, so it is like I am calling

gsub!(args[0]) {nil}

Which is obviously not what I want.  I would like to pass the original 
code block:

gsub!(args[0]) {$1}


Any ideas?

Lou
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