I am trying to define a record #a and a record #b such that #b extends #a, so that I can treat #b (and other subtypes of #a) as #a in some situations. The compiler, though, does not like it and keeps trying to guess the record type based on the first access I开发者_如何学运维 make to it. The following code triggers the warning.
-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]).
Is there anyway I can use this subtyping and make the compilation warning go away? Thanks.
I don't think this can work, because "-record(a,{alfa,beta})." is sort of a template for "{a, alfa, beta}" and "-record(b,{alfa,beta,gama})." results in a tuple "{b,alfa,beta,gama}".
Have a look at http://erlang.org/doc/getting_started/record_macros.html#id66845 , please...
If you want inheritance in Erlang use module inheritance:
-module(inh).
-extends(base).
Also useful "OOP" feature in Erlang is parameterized modules:
-module(param, [Id, Name]).
-compile(export_all).
id() -> Id.
name() -> Name.
and use it:
P = param:new(1, stas).
P:id(). % returns 1
P:name(). % returns stas
Maybe combination of these features helps you.
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