Ruby等效于python nonlocal

Question

I am trying to write a closure in Ruby. This is the code written in Python:

def counter():
    x = 0
    def increment(y):
        nonlocal x
        x += y
        print(x)
    return increment

Is there a "nonlocal" equivalent in Ruby so I can access and make changes to the variable x from inside increment?

Answer 1

Maybe something like:

class CGroup
  def counter
    @x ||= 0
    lambda do |y|
      @x += y
    end
  end
end

Then:

group = CGroup.new
c = group.counter
c.call(1)
=> 1
c.call(1)
=> 2

I'm not aware of a direct analog for Python's nonlocal .

EDIT : The instance variable is unnecessary, and the same thing can be achieved with a variable local to the method. This makes the class superfluous, although much in Ruby takes place within the context of an object.

Answer 2

Since there is objection to using an object, why not just use a lambda?

counter_generator = ->(){
  x ||= 0
  ->(y){
    x += y
    puts x
  }
}

i = counter_generator.call
=> #<Proc:0x00000100867508@(irb):17 (lambda)>
i.call(1)
1
=> nil
i.call(1)
2
=> nil

Note that the incrementor actually returns nil because you've only specified to output the value of x, not to return it.

Answer 3

The nonlocal keyword tells Python which variables to capture. In Ruby, you don't need such a keyword: all variables are captured unless explicitly mentioned otherwise.

So, the Ruby equivalent to your Python code translates almost directly:

counter = -> {
  x = 0
  ->y {
    x += y
    puts x
  }
}

i = counter.()

i.(2)
# 2

i.(3)
# 5

It would probably be more idiomatic to use a method for counter , though:

def counter
  x = 0
  ->y {
    x += y
    puts x
  }
end

i = counter

i.(2)
# 2

i.(3)
# 5