Erlang:主管:start_child / 2错误令我感到困惑

时间:2013-06-05 19:53:26

标签: erlang otp supervisor gen-server

我正在对Logan / Merritt / Carlson的简单缓存,第6章,第149-169页,Erlang和OTP in Action进行略微修改。到目前为止,没有代码更改,只需重命名模块。

我开始申请:

application:start(gridz).
ok

我插入一个项目:

gridz_maker:insert(blip, blop).

我收到此错误:

** exception error: no match of right hand side value 
                {error,
                    {function_clause,
                        [{gridz_edit,init,
                             [{blop,86400}],
                             [{file,"src/gridz_edit.erl"},{line,51}]},
                         {gen_server,init_it,6,
                             [{file,"gen_server.erl"},{line,304}]},
                         {proc_lib,init_p_do_apply,3,
                             [{file,"proc_lib.erl"},{line,227}]}]}}
 in function  gridz_maker:insert/2 (src/gridz_maker.erl, line 15)

以下是代码:

insert(Key, Value) ->
   case gridz_store:lookup(Key) of
      {ok, Pid}  -> gridz_edit:replace(Pid, Value);
      {error, _} -> {ok, Pid} = gridz_edit:create(Value),   %% line 15
                    gridz_store:insert(Key, Pid)
   end.

我看第15行:

  {error, _} -> {ok, Pid} = gridz_edit:create(Value),

我期待错误,因为这是一个新项目。 gridz:edit是gen_server(Logan等人的sc_element)这是create / 1的代码:

create(Value) ->
  create(Value, ?DEFAULT_LEASE_TIME).

create(Value, LeaseTime) ->
   gridz_sup:start_child(Value, LeaseTime).

这是gridz_sup的代码:start_child / 2:

start_child(Value, LeaseTime) ->
   supervisor:start_child(?SERVER, [Value, LeaseTime]).

init([]) ->
   Grid            = {gridz_edit, {gridz_edit, start_link, []},
                     temporary, brutal_kill, worker, [gridz_edit]},
   Children        =  [Grid],
   RestartStrategy = {simple_one_for_one, 0, 1},
                     {ok, {RestartStrategy, Children}}.

如果我直接执行supervisor:start_child / 2,这就是我得到的:

{error,{function_clause,[{gridz_edit,init,
                                 [{blop,50400}],
                                 [{file,"src/gridz_edit.erl"},{line,51}]},
                     {gen_server,init_it,6,
                                 [{file,"gen_server.erl"},{line,304}]},
                     {proc_lib,init_p_do_apply,3,
                               [{file,"proc_lib.erl"},{line,227}]}]}}

gridz_edit中的第51行是init函数:

init([Value, LeaseTime]) ->
   Now = calendar:local_time(),
   StartTime = calendar:datetime_to_gregorian_seconds(Now),
   {ok,
   #state{value = Value,
          lease_time = LeaseTime,
          start_time = StartTime},
   time_left(StartTime, LeaseTime)}.

如果我直接执行它,它可以工作:

120> gridz_edit:init([blop, (60 * 60 * 24)]).
{ok,{state,blop,86400,63537666408},86400000}

所以现在我感到困惑。我错过了什么?为什么supervisor:start_child / 2会抛出错误?

谢谢,

LRP

1 个答案:

答案 0 :(得分:1)

错误说你正在传递一个包含2个成员的元组{blop,86400},当你似乎期待一个包含2个成员的列表时:[Value, LeaseTime]。在直接执行中,您也使用列表,因此它可以正常工作。你应该找出创建元组的位置,然后创建一个列表。