如何在保护子句中使用 'in' 运算符？

Question

我正在尝试在 Elixir 中编写一个字谜检查器。它需要 2 个单词，第一个是参考，第二个是作为第一个的可能变位词进行测试。

我正在尝试用递归和模式匹配来编写它。我在保护子句中使用 in 运算符时出错：

(ArgumentError) invalid args for operator in, it expects a compile time list or range on the right side when used in guard expressions

我不知道如何修复它。这是代码（错误在第 4 个定义中）：

defmodule MyAnagram do
  def anagram?([], []), do: true

  def anagram?([], word) do
    IO.puts 'Not an anagram, the reference word does not contain enough letters'
    false
  end

  def anagram?(reference, []) do
    IO.puts 'Not an anagram, some letters remain in the reference word'
    false
  end

  def anagram?(reference, [head | tail]) when head in reference do
    anagram?(reference - head, tail)
  end

  def anagram?(_, [head | _]) do
    IO.puts 'Not an anagram, #{head} is not in the reference word.'
    false
  end
end

Answer 1

这是由以下代码（如您所识别）引起的：

def anagram?(reference, [head | tail]) when head in reference do
  anagram?(reference - head, tail)
end

您可以找到 in 宏 in the source code 的定义，但为了方便起见，我将其复制到此处 - 它在文档中还包含以下内容：

Guards

The in operator can be used in guard clauses as long as the right-hand side is a range or a list. In such cases, Elixir will expand the operator to a valid guard expression. For example:
  when x in [1, 2, 3] 
translates to:
  when x === 1 or x === 2 or x === 3

定义宏的代码：

  defmacro left in right do
    in_module? = (__CALLER__.context == nil)

    right = case bootstraped?(Macro) and not in_module? do
      true  -> Macro.expand(right, __CALLER__)
      false -> right
    end

    case right do
      _ when in_module? ->
        quote do: Elixir.Enum.member?(unquote(right), unquote(left))
      [] ->
        false
      [h|t] ->
        :lists.foldr(fn x, acc ->
          quote do
            unquote(comp(left, x)) or unquote(acc)
          end
        end, comp(left, h), t)
      {:%{}, [], [__struct__: Elixir.Range, first: first, last: last]} ->
        in_range(left, Macro.expand(first, __CALLER__), Macro.expand(last, __CALLER__))
      _ ->
        raise ArgumentError, <<"invalid args for operator in, it expects a compile time list ",
                                        "or range on the right side when used in guard expressions, got: ",
                                        Macro.to_string(right) :: binary>>
    end
  end

您的代码块到达了 case 语句的最后一部分，因为在编译时无法保证您的变量 reference 的类型为 list（或 range .)

您可以通过调用查看传递给宏的值：

iex(2)> quote do: head in reference                                       
{:in, [context: Elixir, import: Kernel],
 [{:head, [], Elixir}, {:reference, [], Elixir}]}

在这里，原子 :reference 被传递给 in 宏，它与前面的任何子句都不匹配，因此它落入 _ 子句（引发错误。）

要解决此问题，您需要将最后两个子句合并到一个函数中：

  def anagram?(reference, [head | tail]) do
    case head in reference do
      false ->
        IO.puts 'Not an anagram, #{head} is not in the reference word.'
        false
      true ->
        anagram?(reference - head, tail)
    end
  end

还值得注意的是，您可能想使用 "strings" 而不是 'char_lists' http://elixir-lang.org/getting-started/binaries-strings-and-char-lists.html#char-lists

另一件事是调用 reference - head 将不起作用（它会引发 ArithmeticError）。您可能需要查看 List.delete/2 以从列表中删除项目。

如何在保护子句中使用 'in' 运算符？

How to use the 'in' operator in guard clauses?

pattern-matching

elixir

anagram

Guards