doctest可以使用默认参数解析函数吗？

Question

对于我编写的函数，doctest失败。我得到的错误消息是generate_operands/0未定义。我的意图是在文档测试中使用generate_operands/6的默认参数--这可能吗？做我想做的事情是明智的，还是错误是故意的？

该函数如下所示：

@doc """
Generate the operands for arithmetic practice problems.

## Examples
    iex> operands = generate_operands()
    iex> length(operands)
    10
    iex> hd(operands) |> length()
    5
"""
@spec generate_operands(
        integer,
        integer,
        boolean,
        integer,
        boolean,
        SimplePRNG.t()
      ) :: [
        [integer()],
        ...
      ]

def generate_operands(
      num_probs \\ 10,
      num_ops \\ 5,
      equal_num \\ false,
      max_power \\ 5,
      equal_mag \\ false,
      prng \\ SimplePRNGImpl
    )

def generate_operands(num_probs, num_ops, equal_num, max_power, equal_mag, prng) do
  IO.puts("generate_operands()")
end

错误消息为：

== Compilation error in file test/arithmetic_practice_test.exs ==
** (CompileError) (for doctest at) lib/arithmetic_practice.ex:25: undefined function generate_operands/0
    (stdlib) lists.erl:1338: :lists.foreach/2
    (stdlib) erl_eval.erl:680: :erl_eval.do_apply/6
    (elixir) lib/code.ex:767: Code.require_file/2
    (elixir) lib/kernel/parallel_compiler.ex:209: anonymous fn/4 in Kernel.ParallelCompiler.spawn_workers/6 

Answer 1

您需要使用到generate_operands的带有ModuleName的完整路径。检查下面的代码。

@doc """
Generate the operands for arithmetic practice problems.

## Examples
iex> operands = MyModule.InnerModule.generate_operands()
iex> length(operands)
10
iex> hd(operands) |> length()
5
"""