Ruby 使用yield更改函数内的参数

我不熟悉收益率的概念，目前正在实践它。 我本想得到[面包，多汁的，面包] 但是我有[面包，牛排，面包]

你能帮我理解为什么吗？我如何修复我的代码

def burger(patty)
  if block_given?
    yield(patty)
  end
  return ["bread", patty, "bread"]
end


# TODO: Change 'steak'to 'JUICY'using yield
juicy_burger = burger("steak") do |patty|
  patty = "JUICY"
end
p juicy_burger

这样的重新分配不会传播到外部作用域。一旦该块返回，patty将恢复到其初始值

解决方案？在覆盖值仍在范围内时使用该值

def burger(patty)
  if block_given?
    patty = yield(patty)
    # ^ shadow the parameter here. 
    # doesn't have to be a shadow. You can use any name. new_patty = yield(patty)
  end
  return ["bread", patty, "bread"]
end


juicy_burger = burger("steak") do |patty|
  "JUICY #{patty}" # <- simple return here
end
p juicy_burger # >> ["bread", "JUICY steak", "bread"]

---

变量是函数的局部变量，您没有保存返回的值。下面的代码将为您提供[面包，多汁，面包]

---

就像sergio提到的，没有必要在块内赋值，因为这是一个很好的答案。只需补充一点，块参数“patty”没有使用或不需要，可以从方法调用和产量中省略。@SteveTurczyn：我的版本中使用了它，但你是对的。

def burger(patty)
  if block_given?
    patty = yield(patty)
    # ^ shadow the parameter here. 
    # doesn't have to be a shadow. You can use any name. new_patty = yield(patty)
  end
  return ["bread", patty, "bread"]
end


juicy_burger = burger("steak") do |patty|
  "JUICY #{patty}" # <- simple return here
end
p juicy_burger # >> ["bread", "JUICY steak", "bread"]

def burger(patty)
  if block_given?
    patty = yield(patty) # <-- this is the diff
  end
  return ["bread", patty, "bread"]
end


# TODO: Change 'steak'to 'JUICY'using yield
juicy_burger = burger("steak") do |patty|
  patty = "JUICY"
end
p juicy_burger