如何使用erlang的编译器批准子类型记录？

Question

我正在尝试定义记录#a和记录#b,以便#b扩展#a,这样我就可以在某些情况下将#b(以及#a的其他子类型)视为#a.但是编译器并不喜欢它,并且一直试图根据我对它的第一次访问来猜测记录类型.以下代码触发警告.

-module(sandbox).  
-record(a,{alfa,beta}).  
-record(b,{alfa,beta,gama}).  
-export([test/0]).
test() ->  
    A = #b{alfa=1,beta = 2,gama=3},  
    self()!{msg,A},  
    receive  
    {msg,Msg} ->  
        Alfa = Msg#b.alfa,
        Beta = Msg#b.beta,
        case is_record(Msg,b) of
            true ->
                Gama = Msg#b.gama;
            false->                  %% Warning. Erlang assumes that Msg is a #b and therefore this will never match.
                Gama = []
        end
end,
io:format("~p ~p ~p",[Alfa,Beta,Gama]).


test1() ->
A = #b{alfa=1,beta = 2,gama=3},
self()!{msg,A},
receive
    {msg,Msg} ->
        Alfa = Msg#a.alfa,
        Beta = Msg#a.beta,
        case is_record(Msg,b) of
            true ->                  %% Warning. Erlang assumes that Msg is an #a, and therefore this will never match.
                Gama = Msg#b.gama;
            false->
                Gama = []
        end
end,
io:format("~p ~p ~p",[Alfa,Beta,Gama]).

无论如何我可以使用这个子类型并使编译警告消失吗？谢谢.

Answer 1

我不认为这可行,因为"-record(a,{alfa,beta})." 是"{a,alfa,beta}"和"-record(b,{alfa,beta,gama})的模板." 结果是一个元组"{b,alfa,beta,gama}".

请查看http://erlang.org/doc/getting_started/record_macros.html#id66845,请...