4

有没有更惯用的方法来对记录进行模式匹配？我的代码似乎不正确。

type Period = AM | PM

type TimeOfDay = {hours : int; minutes : int; p : Period}

let before (tod1 : TimeOfDay, tod2 : TimeOfDay) =
   match tod1, tod2 with
   | {hours = h1; minutes = m1; p = AM}, {hours = h2; minutes = m2; p = AM} -> (h1, m1) < (h2, m2)
   | {hours = h1; minutes = m1; p = PM}, {hours = h2; minutes = m2; p = PM} -> (h1, m1) < (h2, m2)
   | {hours = _; minutes = _; p = AM}, {hours = _; minutes = _; p = PM} -> true
   | {hours = _; minutes = _; p = PM}, {hours = _; minutes = _; p = AM} -> false
4

2 回答 2

6

您可以稍作改进，因为您不需要显示不需要的模式来生成以下内容

let before (tod1 : TimeOfDay, tod2 : TimeOfDay) =
   match tod1, tod2 with
   | {hours = h1; minutes = m1; p = AM}, {hours = h2; minutes = m2; p = AM} -> (h1, m1) < (h2, m2)
   | {hours = h1; minutes = m1; p = PM}, {hours = h2; minutes = m2; p = PM} -> (h1, m1) < (h2, m2)
   | { p = AM}, {p = PM} -> true
   | { p = PM}, {p = AM} -> false

接下来，您可以定义一个活动模式以将类型解构为元组，如下所示

let (|TIME|) (t:TimeOfDay) = t.hours,t.minutes,t.p

let before (tod1 : TimeOfDay, tod2 : TimeOfDay) =
   match tod1, tod2 with
   | TIME(h1,m1,AM), TIME(h2,m2,PM) -> (h1, m1) < (h2, m2)
   | TIME(h1,m1,PM), TIME(h2,m2,PM) -> (h1, m1) < (h2, m2)
   | { p = AM}, {p = PM} -> true
   | { p = PM}, {p = AM} -> false
于 2013-10-11T10:24:16.220 回答
1

您可以删除匹配中不需要的属性，使用“守卫”（when 子句）检查两个p属性是否相等，并创建一个局部函数来简化元组比较。

let before (tod1: TimeOfDay, tod2: TimeOfDay) =
   let getTime tod = (tod.hours, tod.minutes)
   match tod1, tod2 with
   | { p = x }, { p = y } when x = y -> getTime tod1 < getTime tod2
   | { p = AM }, _ -> true
   | _ -> false

或删除最后一个模式只是做

let before (tod1: TimeOfDay, tod2: TimeOfDay) =
   let getTime tod = (tod.hours, tod.minutes)
   match tod1, tod2 with
   | { p = x }, { p = y } when x = y -> getTime tod1 < getTime tod2
   | { p = x }, _ -> x = AM
于 2020-02-20T03:05:48.373 回答