From: MonkeeSage <MonkeeSage@...>
Date: 2007-12-25T07:49:58+09:00
Subject: Re: Is there any way to pass further the "hidden" block?

On Dec 21, 6:38 am, MonkeeSage <MonkeeS...@gmail.com> wrote:
> On Dec 17, 7:49 pm, MonkeeSage <MonkeeS...@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?
>
> > 2.) Why is it less expensive to pass a Proc as an lval rather than as
> > a block_arg?
>
> > Anybody know?
>
> > Regards,
> > Jordan
>
> Bump

Bump...anyone have any idea about the two questions I asked?