Question

以下折叠功能会将重复的玩家加载到基本记录中：

$scope.validValues = [{
  "code": "LKG",
  "decode": "LKG"
}, {
  "code": "UKG",
  "decode": "UKG"
}, {
  "code": "1",
  "decode": "class 1"
}, {
  "code": "2",
  "decode": "class 2"
}, {
  "code": "3",
  "decode": "class 3"
}, {
  "code": "4",
  "decode": "class 4"
}, {
  "code": "5",
  "decode": "class 5"
}, {
  "code": "6",
  "decode": "class 6"
}, {
  "code": "7",
  "decode": "class 7"
}, {
  "code": "8",
  "decode": "class 8"
}, {
  "code": "9",
  "decode": "class 9"
}, {
  "code": "10",
  "decode": "class 10"
}, {
  "code": "11",
  "decode": "class 11"}
  ];

测试如下：

    (initializedBase, plays) ||> List.fold (fun bases play -> 
                                                Some play.Player |> move bases)

实际结果是：

[<Test>]
let ``2 players each hit single``() =
    // Setup
    let players = [{ Player= Scott; Hit= Single }; { Player= Brian; Hit= Single }]

    // Test
    let bases = players |> assignBases

    bases |> should equal { First=Some Brian; Second=Some Scott; Third=None }

我期待：

{ First=Some Scott; Second=Some Scott; Third=None }

整个代码在这里：

{ First=Some Brian; Second=Some Scott; Third=None }

Answer 1

问题似乎与Seq.fold无关。如果我没弄错的话，错误发生在move函数中匹配表达式的第二行：

| { First=player; Second=None; Third=None } -> { bases with First=player; Second=bases.First }

该模式使用player来捕获First的{{1}}字段的值。虽然看起来应该是这样，但不匹配来自匹配表达式之外的bases的值！如果您需要，则需要另外使用player子句。

实际上，从F＃确定第二个案例是正确的案例（when和Second为Third而None有{First时1}} value），直到控件离开整个匹配表达式，名称Some表示新绑定，并且被称为“遮蔽”旧的player值。 / p>

因此，当您在该匹配案例的结果部分中将player分配给player时，这又将是First的原始值 - 与您为{设置的相同{1}}。

那么你想要的第二个匹配案例是

bases.First

现在，Second仍然引用与匹配表达式之外的值相同的值，即您要在| { First=oldFirst; Second=None; Third=None } -> { bases with First=player; Second=oldFirst }输入字段的值，player获取您的播放器“从“First获取”。

折叠功能不会产生预期的结果

1 个答案: