可变范围和修改

Question

为什么调用mess_with_vars方法不会修改？

中显示的变量的值

def mess_with_vars(one, two, three)
  one = "two"
  two = "three"
  three = "one"
end

one = "one"
two = "two"
three = "three"

mess_with_vars(one, two, three)

puts "one is: #{one}"
puts "two is: #{two}"
puts "three is: #{three}"

Answer 1

Ruby是按值传递（Is Ruby pass by reference or by value?），所以你绝对可以修改对象的值，你会在方法之外看到它的效果。

考虑这些：

def mess_with_vars(one, two, three)
  one.gsub!('one','two')
  two.delete!("wo")
  three.replace "one"
end

以上所有修改参数。

Answer 2

因为在方法定义中初始化的局部变量的范围是方法定义。