Ruby array manipulation inside method

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傲寒
傲寒 2021-01-19 01:05

In the following, input_1 changes:

def method_1(a)
  a << \"new value\"
end

input_1 = []
method_1(input_1)
input_1 #=> [\"new value\"]         


        
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3条回答
  • 2021-01-19 01:32

    Why input_1 changes whereas input_2 doesn't change?

    The very simple answer to this question is that your premise is wrong. input_1 doesn't change. The object that input_1 references changes, but that is something completely different from input_1. The name of a thing is not the same as the thing itself. (Outside of witchcraft.)

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  • 2021-01-19 01:43

    It boils down to Ruby using "pass-reference-by-value".

    The exact case that you encounter is described in this excellent blog post.

    The gist:

    In method_1 you are changing the value of an object that two different variables (input_1 and a) are both pointing to.

    In method_2 you are reassigning a completely new object to one of the two variables (a).

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  • 2021-01-19 01:46

    With a bit of simplification we can say that a variable in Ruby is a reference to a value. In your case variable a holds a reference to an array.

    a << (a.append) mutates the value stored in variable a. The reference is not changed, but the value did. It's the case of method_1

    def method_1(a)
        a << "new value"
    end
    

    Assignment = changes the reference stored in a variable - it starts to point to a different value. References are copied when passed to a method. Because of that when you call

    def method_2(a)
        a = ["new value"]
    end
    input = []
    method_2(a)
    

    You only change a reference stored in a that is local to the method, without any change to the reference stored in input nor to the value (and array of []) that is pointed by this reference.

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