From: Robert Klemme <shortcutter@...>
Date: 2010-10-23T15:50:14+09:00
Subject: Re: Why no yield in 1.9.2 blocks?

On 22.10.2010 22:09, Ed Howland wrote:
> I searched in vain for the answer, but why doesn't yield work inside
> Ruby 192 blocks. The syntax allows you to pass the blocks as a&blk
> param, but you have to call the block with blk.call. If you try to use
> yield, you get a LocalJumpError:

This isn't something new in 1.9.2.  There is no version of Ruby that I 
am aware of that would allow this.

> Why would you want to pass a block to a block? Why not? so you can do
> things like:
>
> def meth(x,&block)
>    if block_given?
>      block.call(x) do |y|
>        puts "in meth " + y.to_s
>      end
>    end
> end
>
> ruby-1.9.2-p0>  meth(1) {|x,&blk| puts "in blok " + x.to_s; blk.call 2}
> in blok 1
> in meth 2

There are also other useful usecases for this, namely defining methods 
with a block dynamically

# in some class
define_method :foo do |x,y,&b|
   puts "before"
   yield # or b.call
   puts "after"
end

Kind regards

	robert


-- 
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/