In the following, input_1 changes:
def method_1(a)
a << "new value"
end
input_1 = []
method_1(input_1)
input_1 #=> ["new value"]
In the following, input_2 does not change:
def method_2(a)
a = ["new value"]
end
input_2 = []
method_2(input_2)
input_2 #=> []
Why does input_1 change while input_2 doesn't change?
It boils down to Ruby using "pass-reference-by-value".
The exact case that you encounter is described in this excellent blog post.
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).
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.
input_2 = method_2(input_2), the new reference generated by method_2 would be assigned to input_2 since Ruby methods return the result of their last evaluation.Why
input_1changes whereasinput_2doesn'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.)
['new value']both times?method_2simply assignes a new variable with the same name as the other one.