实现（Mnesia样式/通配符/不关心）元组匹配（Erlang）的正确方法

我正在执行一个象棋游戏，中国象棋，又名。香旗，确切地说是二郎

工件由{Color，Type}元组表示，点即位置由{File，Rank}元组表示。电路板由点对点映射表示，即{point=>piece}

有一个查询板上某一点是否被工件占用的功能：

is_point_occupied_simple(Board, Point) ->
    ensure_is_point(Point),
    case maps:find(Point, Board) of
        {ok, _} ->
            true;
        error ->
            false
    end.

但是，我想添加一个可选参数来检查工件的颜色-如果该点被指定颜色的工件占据，则函数返回true；否则返回false。如果我不关心工件的颜色，我可以只在TargetColor参数中输入“u”，或者等效地调用is_point_occulated/2：

我不喜欢上面的实现，因为复制和粘贴的比例很大，所以我简化了上面的功能，如下所示：

is_point_occupied_2(Board, Point) ->
    is_point_occupied_2(Board, Point, '_').

is_point_occupied_2(Board, Point, TargetColor) ->
    ensure_is_point(Point),
    ensure_is_color_or_wildcard(TargetColor),
    case maps:find(Point, Board) of
        {ok, {TargetColor, _}} ->
            true;
        {ok, _} ->
            is_wildcard(TargetColor);
        error ->
            false
    end.

函数为\u通配符/1只是一行：

is_wildcard(Wildcard) -> Wildcard =:= '_'.

现在，我想进一步用TargetPiece替换TargetColor，这是一个{TargetColor，TargetType}元组。无、一个或两个元组元素可以是通配符“\u1”。我发现写案例条款很困难。我还注意到，要以这种方式匹配n元组，需要2n子句。因此，显然这不是实现这一目标的正确方式

有人有更好的主意吗

---

PS：我没有包括所有函数的源代码，因为我认为那些我没有包括的函数实现起来很简单。如果您感兴趣，请在下面留下评论。谢谢

我的解决方案是实现一个辅助匹配函数，将上面提到的2n个case子句变成n个orelse子句加上n-1和also子句：

主要功能变化不大：

is_point_occupied_3(Board, Point) ->
    is_point_occupied_3(Board, Point, {'_', '_'}).

is_point_occupied_3(Board, Point, TargetPiece) ->
    ensure_is_point(Point),
    ensure_is_piece_or_wildcard(TargetPiece),
    case maps:find(Point, Board) of
        {ok, Piece} ->
            match_piece(Piece, TargetPiece);
        error ->
            false
    end.

---

有没有更好的/替代的想法？

我会改变棋盘的表示，用一块{颜色，类型}或没有一块{none，none}填充整个棋盘，这样你的代码可以更规则：

-module (testmatch).

-compile([export_all]).

is_point_occupied(Board,Point) -> not match(maps:find(Point, Board),{none,none}).
is_point_occupied(Board,Point,Color) -> match(maps:find(Point, Board),{Color,'_'}).
is_point_occupied(Board,Point,Color,Type) -> match(maps:find(Point, Board),{Color,Type}).

match({ok,{C,T}},{C,T}) -> true;
match({ok,{_C,T}},{'_',T}) -> true;
match({ok,{C,_T}},{C,'_'}) -> true;
match(_,_) -> false.

test() ->
    Board = #{1 =>{none,none}, 2 =>{black,king}, 3 => {white,tower} },
    false = is_point_occupied(Board,1),
    true = is_point_occupied(Board,2),
    false = is_point_occupied(Board,2,red),
    true = is_point_occupied(Board,2,black),
    false = is_point_occupied(Board,3,'_',king),
    true = is_point_occupied(Board,3,'_',tower),
    true = is_point_occupied(Board,2,black,king),
    false = is_point_occupied(Board,2,black,tower),
    false = is_point_occupied(Board,2,white,king).

---

谢谢你的回答！那么，这是否意味着你的棋盘需要8 x 8=64个象棋元素和9 x 10=90个中国象棋元素？由于内存消耗在这里很重要，我认为这可能不是一个好方法。此外，为了匹配n元组{E1，E2，…，En}，必须编写2^n match/2子句。正如我提到的，我怀疑这样做是否正确。顺便说一下，国际象棋在这里只是作为一个例子，我主要关注的是不在乎风格匹配。是否必须更改董事会定义？有没有比我更优雅的解决方案来实现这一点？谢谢

is_point_occupied_3(Board, Point) ->
    is_point_occupied_3(Board, Point, {'_', '_'}).

is_point_occupied_3(Board, Point, TargetPiece) ->
    ensure_is_point(Point),
    ensure_is_piece_or_wildcard(TargetPiece),
    case maps:find(Point, Board) of
        {ok, Piece} ->
            match_piece(Piece, TargetPiece);
        error ->
            false
    end.

-module (testmatch).

-compile([export_all]).

is_point_occupied(Board,Point) -> not match(maps:find(Point, Board),{none,none}).
is_point_occupied(Board,Point,Color) -> match(maps:find(Point, Board),{Color,'_'}).
is_point_occupied(Board,Point,Color,Type) -> match(maps:find(Point, Board),{Color,Type}).

match({ok,{C,T}},{C,T}) -> true;
match({ok,{_C,T}},{'_',T}) -> true;
match({ok,{C,_T}},{C,'_'}) -> true;
match(_,_) -> false.

test() ->
    Board = #{1 =>{none,none}, 2 =>{black,king}, 3 => {white,tower} },
    false = is_point_occupied(Board,1),
    true = is_point_occupied(Board,2),
    false = is_point_occupied(Board,2,red),
    true = is_point_occupied(Board,2,black),
    false = is_point_occupied(Board,3,'_',king),
    true = is_point_occupied(Board,3,'_',tower),
    true = is_point_occupied(Board,2,black,king),
    false = is_point_occupied(Board,2,black,tower),
    false = is_point_occupied(Board,2,white,king).