Question

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，我想制作一个新的辣酱玉米饼馅就像它除了subsubsubrecords之一的值。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}.

只是思考这是痛苦的。还有更好的方法吗？

Answer 1

正如Hynek建议的那样，你可以忽略临时变量并执行：

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

Yariv Sadan对同一问题感到沮丧并撰写了Recless，a type inferring parse transform for records，可以让您写下来：

-compile({parse_transform, recless}).

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

Answer 2

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，结构不是Erlang最强大的部分。

有没有更简单的方法来修改Erlang中subsubsub记录字段中的值？

2 个答案: