From: "David A. Black" <dblack@...>
Date: 2004-12-08T04:19:11+09:00
Subject: Re: Using yield

Hi --

On Tue, 7 Dec 2004, Robert Klemme wrote:

> 13:21:36 [robert.klemme]: ruby yield-vs-block.rb
>                           user     system      total        real
> yield                 0.250000   0.000000   0.250000 (  0.249000)
> block                 0.125000   0.000000   0.125000 (  0.124000)
> 13:21:44 [robert.klemme]: cat yield-vs-block.rb
> 
> require 'benchmark'
> include Benchmark
> 
> REP = 100000
> 
> def bl_yield(n)
>   n.times {|i| yield i}
> end
> 
> def bl_fwd(n,&b)
>   n.times(&b)
> end

Those aren't equivalent, though.  The second one would have to be:

  n.times {|i| b.call(i) }

to compare the two idioms directly.  

When I make that change I get:

yield                 0.320000   0.000000   0.320000 (  0.332194)
block                 0.460000   0.000000   0.460000 (  0.490299)


> I blieve the yield forwarding is slower because there is one more call.
> 
> > Wouldn't the &block form be more clutter?  Well, not that it's so
> > much clutter in either case, but I think it would be slightly wordier
> > (def five_time(&block) etc.)
> 
> For forwarding I prefer &b because the original block is simply passed on
> vs. a new block yields to the original block.  That seems awkward to me
> and seems to be slower, too.  Personally I find the yield forwarding more
> clutter.  YMMV

The clutter point was (at least in my mind :-) strictly about the
difference between:

   def x
     yield
   end

and

   def x(&b)
     b.call
   end

As for forwarding a block to a new method -- isn't &b the only way?
Or am I behind the times on automatic propagation?  (But even in that
case I don't think supplying block #1 to method #2 is the same as
yielding to block #1 during method #1.)  


David

-- 
David A. Black
dblack@wobblini.net