list - Erlang 记录项列表

Question

例如我有erlang记录:

-record(state, {clients
            }).

我可以从客户字段列表中制作吗？

我可以像正常列表一样保留在客户中吗？我怎样才能在这个列表中添加一些值？

谢谢你。

Answer 1

也许你的意思是:

-module(reclist).
-export([empty_state/0, some_state/0, 
         add_client/1, del_client/1,
         get_clients/1]).

-record(state, 
     {    
          clients = []   ::[pos_integer()],
          dbname         ::char()
     }).    

empty_state() ->
     #state{}.

some_state() ->
     #state{
          clients = [1,2,3],
          dbname  = "QA"}.

del_client(Client) ->
     S = some_state(),
     C = S#state.clients,
     S#state{clients = lists:delete(Client, C)}. 

add_client(Client) ->
     S = some_state(),
     C = S#state.clients,
     S#state{clients = [Client|C]}.

get_clients(#state{clients = C, dbname = _D}) ->
     C.

测试:

1> reclist:empty_state().
{state,[],undefined}
2> reclist:some_state(). 
{state,[1,2,3],"QA"}
3> reclist:add_client(4).
{state,[4,1,2,3],"QA"}
4> reclist:del_client(2).
{state,[1,3],"QA"}

::[pos_integer()]表示字段的类型是一个正整数值列表，从 1开始;这是分析工具的提示 dialyzer ，当它执行类型检查时。

Erlang 还允许您在记录上使用模式匹配:

5> reclist:get_clients(reclist:some_state()).
[1,2,3]

进一步阅读:

Records

Types and Function Specifications

dialyzer(1)

@只是我的正确意见 answer让我记得我喜欢 Haskell 如何获取数据类型中字段的值。

这是一个数据类型的定义，从 Learn You a Haskell for Great Good! 偷来的。 ，它利用记录语法:

data Car = Car {company :: String 
               ,model   :: String
               ,year    :: Int
               } deriving (Show)

它创建函数 company , model和 year ，即在数据类型中查找字段。我们先造一辆新车:

ghci> Car "Toyota" "Supra" 2005
Car {company = "Toyota", model = "Supra", year = 2005}

或者，使用记录语法(字段的顺序无关紧要):

ghci> Car {model = "Supra", year = 2005, company = "Toyota"}
Car {company = "Toyota", model = "Supra", year = 2005}
ghci> let supra = Car {model = "Supra", year = 2005, company = "Toyota"}
ghci> year supra
2005

我们甚至可以使用模式匹配:

ghci> let (Car {company = c, model = m, year = y}) = supra
ghci> "This " ++ c ++ " " ++ m ++ " was made in " ++ show y
"This Toyota Supra was made in 2005"

我记得有人尝试在 Erlang 中实现类似于 Haskell 的记录语法的东西，但不确定它们是否成功。

一些关于这些尝试的帖子:

In Response to "What Sucks About Erlang"

Geeking out with Lisp Flavoured Erlang .但是我会在这里忽略参数化模块。

当您想要创建某个结构的新值时，LFE 似乎使用了类似于提供 Scheme 的宏(例如 Racket):

> (define-struct car (company model year))
> (define supra (make-car "Toyota" "Supra" 2005))
> (car-model supra)
"Supra"

我希望我们将来会有一些接近 Haskell 记录语法的东西，这将非常实用和方便。