2012-04-26 60 views
0

我写一些代码来测试simple_one_for_one主管,但它不能正常工作,代码:simple_one_for_one孩子不能启动

-module(test_simple_one_for_one). 

-behaviour(supervisor). 

%% API 
-export([start_link/0, start_fun_test/0]). 

%% Supervisor callbacks 
-export([init/1]). 

-define(SERVER, ?MODULE). 

%%-------------------------------------------------------------------- 
start_link() -> 
    {ok, Pid} = supervisor:start_link({local, ?SERVER}, ?MODULE, []). 

start_fun_test() -> 
    supervisor:start_child(test_simple_one_for_one, []). 

init([]) -> 
    RestartStrategy = simple_one_for_one, 
    MaxRestarts = 1000, 
    MaxSecondsBetweenRestarts = 3600, 

    SupFlags = {RestartStrategy, MaxRestarts, MaxSecondsBetweenRestarts}, 

    Restart = permanent, 
    Shutdown = 2000, 
    Type = worker, 

    AChild = {fun_test_sup, {fun_test, run, []}, 
      Restart, Shutdown, Type, [fun_test]}, 
    io:format("start supervisor------ ~n"), 
    {ok, {SupFlags, [AChild]}}. 

当我运行

test_simple_one_for_one:start_link().

test_simple_one_for_one:start_fun_test(). 

in erl shell,它给我错误:

test_simple_one_for_one:start_fun_test(). ** exception exit: {noproc,{gen_server,call, [test_simple_one_for_one,{start_child,[]},infinity]}} in function gen_server:call/3 (gen_server.erl, line 188)

+0

代码似乎没问题,并且在start_link和start_fun_test被调用时工作得很好。这个错误告诉你在start_link之前运行start_fun_test。 – KrHubert 2012-04-26 12:15:37

回答

1

如果这是您为测试编写的所有代码,请注意,当您注册一个主管孩子时,您会提供一个{M,F,A}元组,代表您在启动孩子时调用的函数。

在你的情况下,我认为它不能简单地找到fun_test:run/1函数。