From: MonkeeSage <MonkeeSage@...>
Date: 2007-12-18T10:53:30+09:00
Subject: Re: Is there any way to pass further the "hidden" block?

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