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
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.
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
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.
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.
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