compilation error: record replacement of macro in Erlang - compilation

Question

This is the directory structure.
src/
animal.hrl
people.hrl
data_animal.erl
data_people.erl
test.erl
test_macro.erl
animal.hrl
%% The record definition of animal.
-ifndef(ANIMAL).
-define(ANIMAL,true).
-record(animal,{
id,
animal_name,
age
}).
-endif.
people.hrl
%% The record definition of people.
-ifndef(PEOPLE).
-define(PEOPLE,true).
-record(people,{
id,
people_name,
age
}).
-endif.
data_animal.erl
%% The data file of animal.
-module(data_animal).
-include("animal.hrl").
%% API
-export([get/1,get_ids/0]).
get(1)->
#animal{
id=1,
animal_name="cat",
age=23
};
get(2)->
#animal{
id=2,
animal_name="dog",
age=19
};
get(3)->
#animal{
id=3,
animal_name="tiger",
age=23
};
get(4)->
#animal{
id=4,
animal_name="pig",
age=19
};
get(_)->
undefined.
get_ids()->
[1,2,3,4].
data_people.erl
%% The data file of people.
-module(data_people).
-include("people.hrl").
%% API
-export([get/1,get_ids/0]).
get(1)->
#people{
id=1,
people_name="John",
age=23
};
get(2)->
#people{
id=2,
people_name="Ken",
age=19
};
get(3)->
#people{
id=3,
people_name="Tom",
age=23
};
get(4)->
#people{
id=4,
people_name="Healthy",
age=19
};
get(_)->
undefined.
get_ids()->
[1,2,3,4].
Notice that, for data_animal.erl and data_people.erl, the parameter of get/1is the record's id of the return value, and the return value of get_ids/0 is a list of get/1's parameters.
test.erl
-module(test).
%% API
-export([get_animal_list/1,get_people_list/1]).
-include("animal.hrl").
-include("people.hrl").
get_animal_list(Age)->
Fun=fun(Id,Acc)->
case data_animal:get(Id) of
#animal{age=Age}=Conf->
[Conf|Acc];
_->
Acc
end
end,
lists:foldl(Fun,[],data_animal:get_ids()).
get_people_list(Age)->
Fun=fun(Id,Acc)->
case data_people:get(Id) of
#people{age=Age}=Conf->
[Conf|Acc];
_->
Acc
end
end,
lists:foldl(Fun,[],data_people:get_ids()).
I want to get the data of animal and people, whose ages are 23, so I write 2 functions, get_animal_list/1, get_people_list/1.
I run
1> c(data_animal),c(data_people),c(test).
{ok,test}
2> test:get_people_list(23).
[{people,3,"Tom",23},{people,1,"John",23}]
3> test:get_animal_list(23).
[{animal,3,"tiger",23},{animal,1,"cat",23}]
Suddenly, I find that the 2 functions share the same pattern. Then I attempt to write a macro get_list, and make 2 calls instead.
test_macro.erl
-module(test_macro).
%% API
-export([get_animal_list/1,get_people_list/1]).
-include("animal.hrl").
-include("people.hrl").
-define(get_list(DataMod,Record,Age),(
Fun=fun(Id,Acc)->
case DataMod:get(Id) of
#Record{age=Age}=Conf->
[Conf|Acc];
_->
Acc
end
end,
lists:foldl(Fun,[],DataMod:get_ids())
)).
get_animal_list(Age)->
?get_list(data_animal,animal,Age).
get_people_list(Age)->
?get_list(data_people,people,Age).
But I got the compile error:
4> c(test_macro).
test_macro.erl:22: syntax error before: ','
test_macro.erl:25: syntax error before: ','
test_macro.erl:4: function get_animal_list/1 undefined
test_macro.erl:4: function get_people_list/1 undefined
error
Tell me why~y~y~
Thank you all!
I have 3 questions now.
Is my code really not Erlang-like? It's extracted from my company's project. Am I still thinking in OOP? Or so do the programming guys in my company?
Thanks to #mlambrichs 's advice. It works, but I still wonder why my code get the compilation error? Is it because Erlang preprocessor is a one-pass scanner, so it fails to recognize #Record{age=Age}?
According to #mlambrichs 's suggestion, I try to change the macro
-define(get_list(DataMod, Record, Age),
[P || P <- lists:map(fun(Id) -> DataMod:get(Id) end,
DataMod:get_ids()),
P#Record.age =:= Age]
).
into a function
get_list(DataMod, Record, Age)->
[P || P <- lists:map(fun(Id) -> DataMod:get(Id) end,
DataMod:get_ids()),
P#Record.age =:= Age].
Then I get the compilation error:
syntax error before: Record

Answer 1

The cause of the error is a misplaced '(' which should be removed:
-define(get_list(DataMod,Record,Age), (
^^^
Fun=fun(Id,Acc)->
case DataMod:get(Id) of
#Record{age=Age}=Conf->
[Conf|Acc];
_->
Acc
end
end,
lists:foldl(Fun,[],DataMod:get_ids())
).
EDIT
You added some questions which I would like to start answering now.
Is my code really not Erlang-like?
Usage of macros. There's no real need to use macros in your situation.
In general: you would like to hide the fact what kind of records are used in people and animals. That's implementation and should be shielded by your interface. You can just define a getter function that takes care of that in the right module. Pls. read my rewrite suggestions.
It works, but I still wonder why my code get the compilation error? See top of answer.
....I try to change the macro .... You're right, that function doesn't compile. Apparently a rewrite is needed.
Like this:
get_list(DataMod, Age) ->
[ P || P = {_,_,_,A} <- lists:map(fun(Id) -> DataMod:get(Id) end,
DataMod:get_ids()),
A =:= Age].
EDIT
Taking up a rewrite. What you want is a concatenation of two list in function test (yours test/0, mine test/1). Using a comma doesn't do that for you. ;-)
test(X)->
?get_list(data_animal,X) ++
?get_list(data_people,X).
Let's fix that get_list macro as well. Your macro definition get_list has 3 parameters, where it only needs 2. Why use Record as a parameter when you already use that in get_people_list/1 and get_animal_list/1? For example, try this:
-define(get_list(DataMod, Y),
case DataMod of
data_animal -> get_animal_list(Y);
data_people -> get_people_list(Y);
_Else -> []
end
)
Overall, there is a lot of code replication in your test module. As a follow up to #yjcdll's advice, move the interface functions to animal and people to their own modules.
Let's have a look at your data modules and their get functions, as well.
I would suggest putting all people records in an array, in your case in the data_people module.
people() -> [
#people{ id=1, people_name="John", age=23 },
#people{ id=2, people_name="Ken", age=19 },
#people{ id=3, people_name="Tom", age=23 },
#people{ id=4, people_name="Healthy", age=19 } ].
Next, you would need a getter function to get only the people with a certain age:
get(Age) ->
[X || X <- people(), is_age( X, Age )].
And the is_age/2 function would be:
is_age( Person, Age ) ->
Person#people.age =:= Age.
So in module test your get_people_list/1 would get a lot simpler.
get_people_list(Age) ->
data_people:get(Age).
And so on. Always be on the lookout for code that looks pretty much the same like code you've already used somewhere. Just try to behave as a sane, lazy programmer. Lazy = good. ;-)
EDIT: OP has to stick to modules given. So a rewrite of the macro is:
-define(get_list(DataMod, Record, Age),
[P || P <- lists:map(fun(Id) -> DataMod:get(Id) end,
DataMod:get_ids()),
P#Record.age =:= Age]
).