Question

我试图了解erlang中的流程通信。在这里，我有一个主流程和五个朋友流程。如果朋友向其他5个人中的任何一个发送消息，则他们必须回复。但是主人应该意识到这一切。我正在粘贴下面的代码。

-module(prog).
-import(lists,[append/2,concat/1]).

-import(maps,[from_lists/1,find/2,get/2,update/3]).
-import(string,[equal/2]).
-import(file,[consult/1]).
-export([create_process/1,friends/4, master/1, main/0,prnt/1]).     



%% CREATE PROCESS 
create_process([])->ok;
create_process([H|T])->
    {A,B} = H,
    Pid = spawn(prog,friends,[B,self(),0,A]),
    register(A,Pid),
    create_process(T).


%% FRIENDS PROCESS
friends(Msg, M_pid, State, Self_name)->
    S = lists:concat([Self_name," state =",State,"\n"]),
    io:fwrite(S),
    if
        State == 0 ->
            timer:sleep(500),
            io:fwrite("~p~n",[Self_name]),
            lists:foreach(fun(X) -> whereis(X)!{Self_name,"intro",self()} end, Msg),
            friends(Msg, M_pid, State + 1, Self_name);
        State > 0 ->
            receive
                {Process_name, Process_msg, Process_id} -> 
                    I = equal(Process_msg,"intro"),
                    R = equal(Process_msg,"reply"),
                    XxX = lists:concat([Self_name," recieved ",Process_msg," from ",Process_name,"\n"]),
                    io:fwrite(XxX),
                    if
                        I == true ->
                            io:fwrite("~p~n",[whereis(Process_name)]),
                            M_pid!{lists:concat([Self_name," received intro message from ", Process_name , "[",Process_id,"]"]), self()},
                            io:fwrite(I),
                            whereis(Process_name)!{Self_name, "reply",self()},

                            friends(Msg, M_pid, State + 1, Self_name);
                        R == true ->
                            M_pid!{lists:concat([Self_name," received reply message from ", Process_name , "[",Process_id,"]"]), self()},
                            io:fwrite(R),
                            friends(Msg, M_pid, State + 1, Self_name)
                    end
            after
                1000->
                    io:fwrite(lists:concat([Self_name," has received no calls for 1 second, ending..."]))
            end

    end.


master(State)->
    receive
        {Process_message, Process_id} ->
            io:fwrite(Process_message),
            master(State+1)
    after
        2000->
            ok
    end.


main() ->
    B = [{john, [jill,joe,bob]},
{jill, [bob,joe,bob]},
{sue, [jill,jill,jill,bob,jill]},
{bob, [john]},
{joe, [sue]}],
    create_process(B),
    io:fwrite("~p~n",[whereis(sue)]),

    master(0).

我认为 friends（）函数中的这一行

M_pid!{lists:concat([Self_name," received intro message from ", Process_name , "[",Process_id,"]"]), self()}

是导致错误的原因，但我不明白为什么。 M_pid是已知的，我正在合并所有信息并将其发送给master，但我感到困惑，为什么它不起作用。

我得到的错误如下：

Error in process <0.55.0> with exit value: {function_clause,[{lists,thing_to_list,
                         [<0.54.0>],
                         [{file,"lists.erl"},{line,603}]},
                  {lists,flatmap,2,[{file,"lists.erl"},{line,1250}]},
                  {lists,flatmap,2,[{file,"lists.erl"},{line,1250}]},
                  {prog,friends,4,[{file,"prog.erl"},{line,45}]}]}

我不知道是什么引起了错误。抱歉问了菜鸟问题，谢谢您的帮助。

Answer 1

Dogbert发现的例子：

-module(my).
-compile(export_all).

go() ->
    Pid = spawn(my, nothing, []),
    lists:concat(["hello", Pid]).


nothing() -> nothing.

在外壳中：

2> c(my).
my.erl:2: Warning: export_all flag enabled - all functions will be exported
{ok,my}

3> my:go().
** exception error: no function clause matching 
                    lists:thing_to_list(<0.75.0>) (lists.erl, line 603)
     in function  lists:flatmap/2 (lists.erl, line 1250)
     in call from lists:flatmap/2 (lists.erl, line 1250)
4>

但是：

-module(my).
-compile(export_all).

go() ->
    Pid = spawn(my, nothing, []),
    lists:concat(["hello", pid_to_list(Pid)]).

nothing() -> nothing.

在外壳中：

4> c(my).  
my.erl:2: Warning: export_all flag enabled - all functions will be exported
{ok,my}

5> my:go().
"hello<0.83.0>"

来自erl docs：

concat(Things) -> string()
    Things = [Thing]
    Thing = atom() | integer() | float() | string()

您输入concat（）的列表必须包含原子，整数，浮点数或字符串。 pid既不是原子，整数，浮点数也不是字符串，因此pid不能与concat（）一起使用。但是，pid_to_list()返回一个字符串：

pid_to_list(Pid) -> string()
    Pid = pid()

如您所见，pid有其自己的类型：pid（）。

Answer 2

我运行了您的代码。您出问题的地方是将Process_id（类型为pid()）传递给lists:concat/1。

让我们尝试了解此错误：

{function_clause,[{lists,thing_to_list,
                         [<0.84.0>],
                         [{file,"lists.erl"},{line,603}]},
                  {lists,flatmap,2,[{file,"lists.erl"},{line,1250}]},
                  {lists,flatmap,2,[{file,"lists.erl"},{line,1250}]},
                  {prog,friends,4,[{file,"prog.erl"},{line,39}]}]}

它声明函数lists:thing_to_list/1没有定义（请参见错误日志中的单词function_clause），该函数接受类型pid()的自变量，如[<0.84.0>]所示。

字符串在erlang中表示为列表，这就是为什么我们使用list：concat / 1。

正如@ 7stud所指出的，这些是可以根据documentation传递给lists:concat/1的有效类型：

atom() | integer() | float() | string()

下一行出现2次。修复它们，您就可以开始了：

错误代码：

M_pid!{lists:concat([Self_name," received intro message from ", Process_name , "[",Process_id,"]"]), self()},

更正的代码

M_pid!{lists:concat([Self_name," received intro message from ", Process_name , "[",pid_to_list(Process_id),"]"]), self()},

注意使用功能erlang:pid_to_list/1。根据{{3}}，该函数接受类型pid()并将其返回为string()。

函数子句错误Erlang

2 个答案: