我正在尝试在 Elixir 中编写一个字谜检查器。它需要 2 个单词,第一个是引用,第二个将作为第一个的可能字谜进行测试。
我正在尝试使用递归和模式匹配来编写它。我收到关于使用 in
的错误保护子句中的运算符:
(ArgumentError) invalid args for operator in, it expects a compile time list or range on the right side when used in guard expressions
我不知道该怎么做才能修复它。这是代码(错误在第四个定义中):
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
最佳答案
这是由以下代码(如您所确定的)引起的:
def anagram?(reference, [head | tail]) when head in reference do
anagram?(reference - head, tail)
end
您可以找到
in
的定义。宏 in the source code ,但为方便起见,我将其复制到此处 - 它还包含文档中的以下内容: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
是 list
类型(或 range
。)您可以通过调用以下命令查看传递给宏的值:
iex(2)> quote do: head in reference
{:in, [context: Elixir, import: Kernel],
[{:head, [], Elixir}, {:reference, [], Elixir}]}
在这里,原子
:reference
正在传递给 in
宏,它与前面的任何子句都不匹配,因此它属于 _
子句(引发错误。)要解决此问题,您需要将最后两个子句组合成一个函数:
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
还值得注意的是,您可能想要使用
"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
)。您可能想查看 List.delete/2从列表中删除项目。
关于pattern-matching - 如何在保护条款中使用 'in' 运算符?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31185006/