From: Rick DeNatale <rick.denatale@...>
Date: 2008-01-14T21:58:08+09:00
Subject: Re: Is there any way to pass further the "hidden" block?

On 12/17/07, MonkeeSage <MonkeeSage@gmail.com> wrote:
> Hmmm...another thing...passing prc to #explicit_blk as an lval rather
> than invoking block_pass shaves another second off the time...
>
> def explicit_blk(blk)
>   blk.call("baz")
> end
> n = 1_000_000
> prc = lambda { | y | y }
> Benchmark.bm(10) { | x |
>   x.report("explicit") { n.times { explicit_blk(prc) } }
> }
>
> # =>
>                 user     system      total        real
> explicit    5.030000   0.520000   5.550000 (  5.774020)
>
> ...so I'm curious about two matters:
>
> 1.) Why does creating a Proc explicitly and passing it as a block_arg
> rather than using a literal block (see previous post) speed up the
> #explicit_blk benchmark 2 times, but doesn't have very much effect the
> #implicit_blk benchmark?

because you've moved the expensive proc creation outside of the benchmark loop.
>
> 2.) Why is it less expensive to pass a Proc as an lval rather than as
> a block_arg?

It's not, actually the other way, only slightly:

require "benchmark"

def explicit_blk(blk)
 blk.call("baz")
end
n = 1_000_000
prc = lambda { | y | y }
Benchmark.bm(10) { | x |
  x.report("outside") { n.times { explicit_blk(prc) } }
  x.report("lval") { n.times { prc = lambda { | y | y }; explicit_blk(prc) } }
  x.report("parm") { n.times { explicit_blk(lambda { | y | y }) } }
}

                user     system      total        real
outside     1.430000   0.010000   1.440000 (  1.558680)
lval        5.460000   0.030000   5.490000 (  5.885967)
parm        5.320000   0.030000   5.350000 (  5.748286)



-- 
Rick DeNatale

My blog on Ruby
http://talklikeaduck.denhaven2.com/