Enumerator yielder.yield VS Proc.yield

Question

I recently start to read the book "Programming Ruby 1.9&2.0". It shows a trick for explicit enumerator

triangular_numbers = Enumerator.new do |yielder|
number = 0
count = 1
    loop do
        number += count
        count += 1
        yielder.yield number
    end
end
5.times { print triangular_numbers.next, " " }
puts

I wonder why this yielder.yield will leave the loop temporarily and also return the value of number until the next enumerator object created. It seems different than the usual case when a yield inside the loop block. I check the APIdock and find that the source code of Proc.yield() is the same as Proc.call(). For the Yielder object in the Enumerator class, the Yielder has override the yield(). But why the yielder.yield will temporarily leave the loop block?

Reference: APIdock Yielder yield(), Ruby MRI rb_proc_call

See here: http://stackoverflow.com/questions/993026/help-understanding-yield-and-enumerators-in-ruby/18179273#18179273 If you have any questions, let me know. — 7stud, Sep 18 '13 at 07:38
I think you might be confused about Enumerators. You say "until the next enumerator object [is] created", but there is only one enumerator here: `triangular_numbers`. The enumerator acts as a closure, which is why it can remember the value of `number` between calls. — Max, Sep 18 '13 at 13:54

score 9 · Accepted Answer · edited Oct 29 '16 at 15:16

You are confusing Ruby's yield statement with Enumerator::Yielder's yield method and Proc's yield method. They may be spelled the same but they are completely different.

Statement

The yield statement has no receiver. Inside a method it means "Run the block right now". An error occurs if no block is attached. It is not always given an argument, because sometimes you just want to run the block.

def foo
  yield :bar
end
foo # LocalJumpError
foo { |x| puts x } # bar

Enumerator::Yielder

For a yielder, yield is almost always given an argument. That's because it means the same as << which is "The next time someone calls next on me, give them this value".

Enumerator.new { |yielder| yielder.yield 3 }.next # 3
Enumerator.new { |yielder| yielder << 3 }.next # same thing

I think it's a good idea to use << to avoid confusion with the yield statement.

Proc

Procs and lambdas are basically functions. yield here means the same thing as call, which "Just call the function". You can give it an argument or not, depending on how the proc was defined. Nothing fancy here.

proc { |x| puts x }.yield(:bar) # bar
proc { |x| puts x }.call(:bar) # same thing as previous line

I think it's a good idea to use call to avoid confusion with the yield statement.

After reading this I was just tempted to use `yield` instead of call, because it sounds nicer, but beware, method objects don't respond to `yield`. — , Jul 25 '16 at 23:12

Oleksandr Brezitskyy · Answer 2 · 2014-06-14T15:58:44.977

I also stumbled on that example in the book. After some thinking on how the example might work and browsing Ruby doc I found the Fiber class which I think is what is used by Enumerator behind the scene:

http://www.ruby-doc.org/core-2.0/Fiber.html

Fiber concept implements "light weight cooperative concurrency" which is quite interesting, not that difficult to understand and what's more important it's different from other "yields" for calling blocks or dealing with thread controls.

I think Enumerator has a Fiber object inside which it passes to the block. Then looks like every time when you call "next" on your Enumerator, it calls "resume" on the Fiber object to allow it calculate the next number and when the block calls "yield" on the Fiber, the control goes back to the "next" method. And so on.

Here's my version of possible implementation of Enumerator (of course, only that part that is discussed in the book's example):

class MyExplicitEnumerator

  def initialize (&block)
    @yielder = Fiber.new { block.call Fiber }
  end

  def next
    @yielder.resume
  end

end

e = MyExplicitEnumerator.new do |yielder| 
    number = 1
    loop do
      yielder.yield number
      number += 1
    end
  end

p e.next
p e.next

# output
# 1
# 2

Enumerator yielder.yield VS Proc.yield

2 Answers2

Statement

Enumerator::Yielder

Proc