【问题标题】:Unable to start supervisor: (Protocol.UndefinedError) protocol Enumerable not implemented for无法启动主管:(Protocol.UndefinedError)协议枚举未实现
【发布时间】:2023-03-19 10:01:02
【问题描述】:

我正在尝试启动 Supervisor 和 GenServer:

defmodule SubscriptionManagerSupervisor do
  use Supervisor

  def start_link do
    Supervisor.start_link(__MODULE__, [], [{:name, __MODULE__}])
  end

  # supervisor callback
  def init([]) do
    interval = 1000
    children = worker(SubscriptionManagerServer, [interval])
    supervise(children, strategy: :one_for_one)
  end
end

还有我的 GenServer:

defmodule SubscriptionManagerServer do
  use GenServer
  import Logger

  def start_link(interval) do
    GenServer.start_link(__MODULE__, [interval])
  end

  def init(interval) do
    state = :calendar.universal_time()
    Logger.info "SubscriptionManagerServer init(). State: #{state} interval: #{interval}"
    # This will send message to self on the interval. handle_info handles it.
    :timer.send_interval(interval)
    {:ok, state}
  end

  def handle_info(url, state) do
    new_state = :calendar.universal_time()
    Logger.info "SubscriptionManager handle_info(). new_state: #{new_state}"
    {:noreply, new_state}
  end
end

我尝试开始一切:

defmodule SubscriptionManagerApp do
  use Application
  import Logger

  def start(_type, _args) do
    Logger.info "Starting the SubscriptionManager."
    SubscriptionManagerSupervisor.start_link()
  end
end

但是,我收到以下错误:

** (Mix) Could not start application subscription_manager: SubscriptionManagerApp.start(:normal, []) returned an error: an exception was raised:
    ** c for {SubscriptionManagerServer, {SubscriptionManagerServer, :start_link, [1000]}, :permanent, 5000, :worker, [SubscriptionManagerServer]}
        (elixir) lib/enum.ex:1: Enumerable.impl_for!/1
        (elixir) lib/enum.ex:116: Enumerable.reduce/3
        (elixir) lib/enum.ex:1636: Enum.reduce/3
        (elixir) lib/enum.ex:1188: Enum.map/2
        (elixir) lib/supervisor/spec.ex:169: Supervisor.Spec.supervise/2
        (stdlib) supervisor.erl:294: :supervisor.init/1
        (stdlib) gen_server.erl:328: :gen_server.init_it/6
        (stdlib) proc_lib.erl:247: :proc_lib.init_p_do_apply/3

我尝试更改一些参数并仔细查看文档,但我很难过。

【问题讨论】:

  • 你的GenServer.start_linkinterval 不是[interval]。尝试替换它

标签: elixir erlang-otp erlang-supervisor gen-server


【解决方案1】:

代码有几个问题:

  1. children 需要是一个列表。
  2. GenServer.start_link 的第二个参数应该只是 interval,而不是 [interval]
  3. 在将其注入字符串插值之前,您需要 inspect() 状态,因为它是一个元组。
  4. 没有:timer.send_interval/1,只有/2/3

最终改动:

@@ -8,7 +8,7 @@ defmodule SubscriptionManagerSupervisor do
   # supervisor callback
   def init([]) do
     interval = 1000
-    children = worker(SubscriptionManagerServer, [interval])
+    children = [worker(SubscriptionManagerServer, [interval])]
     supervise(children, strategy: :one_for_one)
   end
 end
@@ -18,20 +18,20 @@ defmodule SubscriptionManagerServer do
   import Logger

   def start_link(interval) do
-    GenServer.start_link(__MODULE__, [interval])
+    GenServer.start_link(__MODULE__, interval)
   end

   def init(interval) do
     state = :calendar.universal_time()
-    Logger.info "SubscriptionManagerServer init(). State: #{state} interval: #{interval}"
+    Logger.info "SubscriptionManagerServer init(). State: #{inspect(state)} interval: #{interval}"
     # This will send message to self on the interval. handle_info handles it.
-    :timer.send_interval(interval)
+    :timer.send_interval(interval, :tick)
     {:ok, state}
   end

   def handle_info(url, state) do
     new_state = :calendar.universal_time()
-    Logger.info "SubscriptionManager handle_info(). new_state: #{new_state}"
+    Logger.info "SubscriptionManager handle_info(). new_state: #{inspect(new_state)}"
     {:noreply, new_state}
   end
 end

演示:

$ mix run --no-halt

16:08:17.771 [info]  Starting the SubscriptionManager.

16:08:17.776 [info]  SubscriptionManagerServer init(). State: {{2016, 9, 7}, {10, 38, 17}} interval: 1000

16:08:18.780 [info]  SubscriptionManager handle_info(). new_state: {{2016, 9, 7}, {10, 38, 18}}

16:08:19.780 [info]  SubscriptionManager handle_info(). new_state: {{2016, 9, 7}, {10, 38, 19}}

16:08:20.784 [info]  SubscriptionManager handle_info(). new_state: {{2016, 9, 7}, {10, 38, 20}}

16:08:21.784 [info]  SubscriptionManager handle_info(). new_state: {{2016, 9, 7}, {10, 38, 21}}

16:08:22.782 [info]  SubscriptionManager handle_info(). new_state: {{2016, 9, 7}, {10, 38, 22}}

^C

【讨论】:

  • 关于主管,我觉得我什么都试过了。但是,我没有意识到孩子必须是一个列表。那是关键。谢谢!
猜你喜欢
  • 2017-11-26
  • 2016-05-09
  • 2016-04-26
  • 2018-05-22
  • 2018-07-28
  • 2015-09-03
  • 1970-01-01
  • 2018-10-19
  • 1970-01-01
相关资源
最近更新 更多