functional-programming - 如何使用 F# 发牌

Question

我一直致力于使用 F# 对流行的纸牌游戏(情书)进行建模，以了解有关函数式编程的更多信息。

module Game = 
open Cards
open Players

type Deck = Card list

let dealACard (deck:Deck) = 
    let randomGenerator = System.Random()
    let index = randomGenerator.Next deck.Length
    let card = deck.Item index
    (card, (deck |> List.filter((<>) card)))

let createPlayer playerNumber deck =
    let card, newDeck = dealACard deck
    ({cards=[card]; playerNumber=playerNumber}, newDeck)

我一直做得很好，直到我学会了如何模拟如何画一张卡片。为了测试这一点，我想从牌组中抽出所有的牌。我的程序看起来像这样:

let deck = createDeck
while not deck.IsEmpty do
    let card, newDeck = dealACard deck
    // print the card
    // how do I update the deck?

任何帮助或反馈都会很棒。

Answer 1

F# 列表是 不可变 ，所以如果 deck.IsEmpty开始 false ，它会留下false永远。不过，真的没有理由让事情变得如此复杂。

假设你有一个排序的甲板。让我们仅以三张牌为例，但假设它是一副完整的牌:

let deck =
    [
        { Suit = Hearts; Face = Queen }
        { Suit = Diamonds; Face = King }
        { Suit = Spades; Face = Ace }
    ]

您可以使用随机数生成器轻松打乱牌组:

let r = Random ()
let scrambledDeck = deck |> List.sortBy (fun _ -> r.Next ())

第一次创建 scrambledDeck ，它在 FSI 中可能是这样的:

> let scrambledDeck = deck |> List.sortBy (fun _ -> r.Next ());;

val scrambledDeck : Card list =
  [{Suit = Spades;
    Face = Ace;}; {Suit = Hearts;
                   Face = Queen;}; {Suit = Diamonds;
                                    Face = King;}]

但是如果你再做一次，它可能看起来像这样:

> let scrambledDeck = deck |> List.sortBy (fun _ -> r.Next ());;

val scrambledDeck : Card list =
  [{Suit = Spades;
    Face = Ace;}; {Suit = Diamonds;
                   Face = King;}; {Suit = Hearts;
                                   Face = Queen;}]

现在你有一个打乱的牌组，你可以简单地开始把牌从上面拉下来，例如为了打印它们:

scrambledDeck |> List.iter (printfn "%O")