如何在保护条款中使用'in'运算符？

Question

我想在Elixir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可能的anagram测试.

我试图用递归和模式匹配来编写它.我in在一个保护条款中使用运算符时收到错误:

(ArgumentError)运算符输入无效的args,它在保护表达式中使用时需要右侧的编译时列表或范围

我不知道该怎么做才能解决它.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

这是由以下代码引起的(如您所示):

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

您可以在源代码中找到in宏的定义,但为了方便起见,我将其复制到此处 - 它还在文档中包含以下内容:

逆天

in只要右侧是范围或列表,操作符就可以用在保护条款中.在这种情况下,Elixir会将运算符扩展为有效的保护表达式.例如:

  when x in [1, 2, 3] 

翻译为:

  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

您的代码块正在命中case语句的最后一部分,因为在编译时无法保证您的变量reference是类型list(或range.)

您可以通过调用以下内容查看传递给宏的值:

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

这里,原子:reference被传递给in宏,它与前面的任何子句都不匹配,因此它会落到_子句中(这会引发错误.)

要解决此问题,您需要将最后两个子句合并为一个函数:

  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

值得注意的是,您可能希望使用"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不起作用(它会引发ArithmeticError).您可能希望查看List.delete/2以从列表中删除项目.