3

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 = ""}).

现在我有一个辣酱玉米饼馅，我想做一个和它一样的新辣酱玉米饼馅，除了其中一个子子子记录的值。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}.

光是想想就很痛苦。有没有更好的办法？

4

2 回答 2

4

正如 Hynek 建议的那样，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}}}.

Yariv Sadan对同样的问题感到沮丧，并编写了Recless，这是一种类型推断解析转换的记录，它允许您编写：

-compile({parse_transform, recless}).

update_attitude(Enchilada = #enchilada{}, Attitude) ->
    Enchilada.stamped_packet.packet.cat.attitude = Attitude.
于 2009-02-27T18:33:59.323 回答
1

试试这个：

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
    }}}}.

无论如何，结构并不是 Erlang 最强大的部分。

于 2009-02-27T17:58:15.617 回答