为什么这个 Ruby 方法通过引用传递它的参数-解网

问：

我回答了这个问题，偶然发现了一些奇怪的事情。 Ruby 按值传递其参数，但变量本身是引用。那么为什么第一种方法似乎通过引用传递其参数呢？

require 'set'
require 'benchmark'

def add_item1!(item, list)
  list << item unless list.include?(item)
end

def add_item2(item, list)
  list |= [item]
end

def add_item3(item, list)
  set = Set.new(list)
  set << item
  list = set.to_a
end

array1 = [3,2,1,4]
add_item1!(5, array1)
p array1 # [3, 2, 1, 4, 5]

array2 = [3,2,1,4]
add_item2(5, array2) 
p array2 # [3, 2, 1, 4]

array3 = [3,2,1,4]
add_item3(5, array3)
p array3 # [3, 2, 1, 4]

Ruby 参数引用 Pass-by-Reference Pass-By-Value

def add_item1!(item, list)
  # MUTATES the list object, which is the original object passed
  # (there is no copy/clone/duplication that occurred)
  list << item unless list.include?(item)
end

def add_item2(item, list)
  # this creates a NEW list and re-assigns it to the parameter
  # re-assigning to a local parameter does not affect the caller
  # the original list object is not modified
  list |= [item]
end

def add_item3(item, list)
  set = Set.new(list)
  set << item
  # this creates a NEW list from set and re-assigns it to the parameter
  # re-assigning to a local parameter does not affect the caller
  # the original list object is not modified
  list = set.to_a
end

为什么这个 Ruby 方法通过引用传递它的参数

Why this Ruby method passes it's argument by reference

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