Question

我编写了一个递归函数doMoves，该函数应与列表的每个元素匹配并调用指定的函数。但是，我运行了一个列表[Draw; Draw]，其中Draw函数仅被调用一次。我不确定为什么它与每个元素都不匹配。

我编写了一个递归函数，该函数应该将列表的每一步都放在头上，并将其与一个函数匹配

let officiate (cards:Card list) (moves:Move list) (goal:int) = 
  let mutable heldCards = []
  let mutable deck = cards

  let doNothing = 
    heldCards

  let DrawCard = 
    lazy (
    heldCards<-List.append heldCards [deck.Head]
    List.length heldCards |> printfn "Length of Drawn heldCards: %d"
    deck<-deck.Tail)

  let rec doMoves movs = 
    match movs with
    | [] -> doNothing
    | x::xs -> 
      match x with
        | Draw -> DrawCard.Force()
      doMoves xs
  doMoves moves
  true

let moves = [Draw; Draw]
let cards = [(Jack,Clubs); (Num(8),Diamonds)]
let card = (Jack,Spades)
officiate cards moves 42

我希望输出状态

Length of Drawn heldCards: 1
Length of Drawn heldCards: 2

但是我得到以下信息：

Length of Drawn heldCards: 1

Answer 1

错误是因为DrawCard是一个值，而不是一个函数。要在F＃中编写无参数函数，您需要指定单位作为输入。即

  let drawCard () = //your implementation

我试图用一种更实用的方式来写我认为你正在尝试的事情：

 let officiate (cards:Card list) (moves:Move list) (goal:int) =

   let drawCard (heldCards: Card list) (deck: Card list) =
     match deck with
     | [] -> Error "Empty deck"
     | head::tail -> Ok (List.append heldCards [head], tail)

   let rec doMoves heldCards deck movs =
     printfn "Hand is %d, deck is %d." (List.length heldCards) (List.length deck)
     match movs with
     | [] -> heldCards, deck
     | x::xs ->
       match x with
         | Draw ->
             drawCard heldCards deck
             |> function
             | Ok (h, d) -> doMoves h d xs
             | Error s ->
                 printfn "An error occurred: %s" s
                 heldCards, deck
         | Discard c -> failwith "todo"
   doMoves [] cards moves

 let moves = [Draw; Draw]
 let cards = [(Jack,Clubs); (Num(8),Diamonds)]
 officiate cards moves 42

F＃不会模式匹配列表的所有元素

1 个答案: