From: MonkeeSage <MonkeeSage@...>
Date: 2007-12-21T21:39:58+09:00
Subject: Re: Is there any way to pass further the "hidden" block?

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

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