Append to an outer variable in a closure

Question

I have the following closure:

def func
  def inner_func
    list << 3 # how to append an element to the outer `list`?
  end

  list = []
  inner_func
  list
end

I am not sure how I can append something to list from inner_func, as the above attempt is wrong.

Problem with Ruby is that it _looks_ like it supports nested methods, but it doesn't, as pointed out by Piotr. You could use eg. Procs though, as indicated in http://www.elonflegenheimer.com/2012/07/08/understanding-ruby-closures.html — EdvardM, Jun 26 '15 at 12:45
I understand. Get you point how to use procs in my case? Also, `inner_func` is actually a whole lot bigger, is it right to stuff everything inside a proc? — linkyndy, Jun 26 '15 at 12:48

score 2 · Accepted Answer · edited May 23 '17 at 12:14

2

Use instance methods, like this:

def func
  def inner_func
    @list << 3 # how to append an element to the outer `list`?
  end

  @list = []
  inner_func
  @list
end

However take a look at this - regarding Ruby and nested methods.

Example of a clean workaround:

def func
  list = []
  inner_func list # => [3]
  inner_func list # => [3, 3]
end

def inner_func(list)
  list << 3
end

edited May 23 '17 at 12:14

Community

answered Jun 26 '15 at 12:33

Piotr Kruczek

score 2 · Answer 2 · answered Jun 26 '15 at 13:18

2

Ruby does not have nested methods. It does have lambda's, which happen to be closures.

def func
  list = []
  l = ->{ list << 3}
  l.call
  p list
end

func # => [3]

answered Jun 26 '15 at 13:18

steenslag

Thank you. Unfortunately, my `inner_func` is recursive, how can I use lambdas in that case? – linkyndy Jun 26 '15 at 13:20
2

Lambda's can be recursive:`fac = ->(n){ |n| n==1 ? 1 : n * fac.call(n - 1)}; fac(5).call` – steenslag Jun 26 '15 at 13:54

2 Answers2