What's the difference between these two block passing strategies?

ruby-talk
1

A common complaint I hear about Ruby is (I think, I may quoting
incorrectly or the wrong thing) that "blocks aren't first order object." I
won't pretend to know what that means but, in context, it appears that they
are complaining about how you can't pass blocks around like variables. In
particular, I recall someone having to deal with some kind of UI toolkit
that requied blocks but couldn't pass the blocks up the call stack.
    I was just fiddling around with Ruby and I think I have two ways to do
this and I was wondering if there were any subtle differences between them
or any gatchas that I might be missing.
    Please take a look at them. Here's the first:

def foo
    yield
end

def bar(&lambda)
    foo &lambda
end

bar { puts "Blocks!" }

    ...and here's the second:

def foo
    yield
end

def bar
    foo { yield }
end

bar { puts "Blocks!" }

    ...so, what do you guys think? Are they equivalent? Is there a
performance difference? Are they even doing what I think they're doing?
    Thank you...

2

...so, what do you guys think? Are they equivalent? Is there
a performance difference? Are they even doing what I think
they're doing?

They are not equivalent. Version A has to convert a block into
a Proc, which slows it down. Version B uses two calls instead
of just one, which slows it down.

Which one wins? See below. The difference in speed is
significant.

gegroet,
Erik V. - http://www.erikveen.dds.nl/

···

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$ cat test.rb
require "ev/ruby"

module A
   def self.foo
     yield
   end

   def self.bar(&lambda)
     foo &lambda
   end
end

module B
   def self.foo
     yield
   end

   def self.bar
     foo { yield }
   end
end

def test(mod)
   bm mod do
     100_000.times do
       mod.bar { :OK }
     end
   end
end

10.times do
   test(A)
   test(B)
end

$ ruby test.rb
        CPU ELAPSED COUNT CPU/COUNT LABEL
  18.926000 18.995000 10 1.892600 A
   2.264000 2.265000 10 0.226400 B

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