From: 7stud -- <bbxx789_05ss@...>
Date: 2009-09-10T13:44:30+09:00
Subject: Re: How Are Variables Kept Independent of Each Other Yet Pas

Yossef Mendelssohn wrote:
>
> Mason, you're going to want to make an actual copy of the original
> variable, not simply a new pointer to it (which is what you get when
> you do something like `b = a`). In many cases, calling .dup or .clone
> will work. (As in `b = a.dup` or `b = a.clone`.)
> 
> However, since you have an array of arrays, you're going to need a
> "deep copy". I believe `b = Marshal.load(Marshal.dump(a))` is the
> standard idiom.

Or, if that offends your sensibilities, in your case you can do this:

arr = [
  [1, 2],
  ["a", "b"]
]


arr_copy = []

arr.each do |subarr|
  arr_copy << subarr.dup
end

p arr
p arr_copy

--output:--
[[1, 2], ["a", "b"]]
[[1, 2], ["a", "b"]]


arr[0][1] = "******"

p arr
p arr_copy

--output:--
[[1, "******"], ["a", "b"]]
[[1, 2], ["a", "b"]]

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