在 Erlang 中是否有更简单的方法来修改 subsubsub 记录字段中的值?

StackOverflow https://stackoverflow.com/questions/595247

  •  09-09-2019
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所以我有一个相当深的记录定义层次结构:

-record(cat,            {name = '_',           attitude = '_',}).
-record(mat,            {color = '_',          fabric = '_'}).
-record(packet,         {cat = '_',            mat = '_'}).
-record(stamped_packet, {packet = '_',         timestamp = '_'}).
-record(enchilada,  {stamped_packet = '_', snarky_comment = ""}).

现在我有了一个辣酱玉米饼馅,我想制作一个新的,除了它的一个子账单的价值外。这就是我一直在做的事情。

update_attitude(Ench0, NewState)
  when is_record(Ench0, enchilada)->

    %% Pick the old one apart.
    #enchilada{stamped_packet     = SP0} = Ench0,
    #stamped_packet{packet = PK0} = SP0,
    #packet{cat = Tag0}    = PK0,

    %% Build up the new one.
    Tude1 = Tude0#cat{attitude = NewState},
    PK1 = PK0#packet{cat = Tude1},
    SP1 = SP0#stamped_packet{packet = PK1},

    %% Thank God that's over.
    Ench0#enchilada{stamped_packet = SP1}.

只是 思维 这很痛苦。有没有更好的办法?

有帮助吗?

解决方案

正如 Hynek 所建议的,您可以省略临时变量并执行以下操作:

update_attitude(E = #enchilada{stamped_packet = (P = #packet{cat=C})},
                NewAttitude) ->
    E#enchilada{stamped_packet = P#packet{cat = C#cat{attitude=NewAttitude}}}.

亚里夫·萨丹 对同样的问题感到沮丧并写道 鲁莽, 记录的类型推断解析转换 这将允许你写:

-compile({parse_transform, recless}).

update_attitude(Enchilada = #enchilada{}, Attitude) ->
    Enchilada.stamped_packet.packet.cat.attitude = Attitude.

其他提示

尝试这种情况:

update_attitude(E = #enchilada{
    stamped_packet = (SP = #stamped_packet{
      packet = (P = #packet{
        cat = C
    })})}, NewState) ->
    E#enchilada{
      stamped_packet = SP#stamped_packet{
        packet = P#packet{
          cat = C#cat{
            attitude = NewState
    }}}}.

反正结构是不二郎的最强大的一部分。

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