[erlang-questions] Exporting a record type

[Fernando Benavides]

Loïc will correct me if I'm wrong but the idea is to use records *only* as
an internal representation of your data types.
For instance, let's say in your system you have shoes (for lack of a better
thing). You'll then create a module called shoes.erl:

-module(shoes).

-export([new/2, id/1, color/1, color/2, size/1, size/2]).

-record(shoe, {id = new_id(), color, size}).
-opaque shoe() :: #shoe{}.
-export_type([shoe/0]).

new(Color, Size) -> #shoe{color = Color, size = Size}.
id(#shoe{id = Id}) -> Id.
color(#shoe{color = Color}) -> Color.
color(Shoe, Color) -> Shoe#shoe{color = Color}.
size(#shoe{size = Size}) -> Size.
size(Shoe, Size) -> Shoe#shoe{size = Size}.

new_id() -> implement:it(yourself).

Whenever you need to work with shoes anywhere in your system, you just use
the functions exported by the module. Let's say you have a list of shoes
and you only want those of a particular size:

filter_by_size(Shoes, Size) ->
  [Shoe || Shoe <- Shoes, shoes:size(Shoe) == Size].

Hope this helps.



*Fernando Benavides <http://google.com/profiles/greenmellon>*


On Thu, Jul 9, 2015 at 5:17 PM, Bob Ippolito <bob@REDACTED> wrote:

> On Thu, Jul 9, 2015 at 5:05 PM, <lloyd@REDACTED> wrote:
>
>>
>> In  Loïc's new book, The Erlanger Playbook, he advocates against using
>> including records through *.hrl. (pp 40-41)
>>
>> Instead, he recommends using types:
>>
>> -record(state, {}).
>> -opaque state() :: #state{}.
>>
>
> This looks like the fairly common recommendation of: don't export records.
> Treat them as an opaque data type, and provide functions for working with
> them.
>
>
>> He leaves hanging the question of how to access this record from another
>> module.
>>
>> The Erlang Types and Function Specifications doc suggests that the type
>> can be exported using -export_type(...). But, to my feeble mind, the
>> exposition is less than clear.
>>
>> My case and problem is as follows:
>>
>> -record(book, {
>>    ... yada yada (some seven fields)
>>    }
>>
>> I define the record, then:
>>
>> -opaque book() :: #book{}.
>>
>> -export_type([book/0]).
>>
>> Now, I access a remote db in a different module and want to instantiate a
>> book record, but it's not at all clear to me how to do so.
>>
>
> You don't. You have a function in the module that defines book that can
> create a book record, and other functions to manipulate them.
>
> An example of this style would be the dict module. It uses records
> internally to implement a dictionary data structure, but you don't have
> direct access to the record(s) that it uses. Just functions to work with
> them.
>
> -bob
>
>
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