Noob在这里,我正在尝试使用52张牌的标准牌创建纸牌游戏。这是我想要成为一个函数的代码所以我不必每次都手动写出来。我正在创建5个不同的玩家,所以这将被复制到其他玩家。
func firstPlayerCardShuffle() {
var firstPlayerCard1Num = (Int(arc4random_uniform(UInt32(13)) + 2))
var firstPlayerCard1Suit = suits[Int(arc4random_uniform(UInt32(suits.count)))]
var firstPlayerCard1 = "\(firstPlayerCard1Num) of \(firstPlayerCard1Suit)"
var firstPlayerCard2Num = (Int(arc4random_uniform(UInt32(13)) + 2))
var firstPlayerCard2Suit = suits[Int(arc4random_uniform(UInt32(suits.count)))]
var firstPlayerCard2 = "\(firstPlayerCard2Num) of \(firstPlayerCard2Suit)"
return(firstPlayerCard1,firstPlayerCard2)
}
有人能让我知道我失踪的是什么。
答案 0 :(得分:4)
不能直接回答你的问题,但我认为你会想要这样的事情:
enum Number: String {
case Two = "two"
case Three = "three"
case Four = "four"
case Five = "five"
case Six = "six"
case Seven = "seven"
case Eight = "eight"
case Nine = "nine"
case Ten = "ten"
case Jack = "jack"
case Queen = "queen"
case King = "king"
case Ace = "ace"
static var randomNumber: Number {
return [Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King, Ace][Int(arc4random_uniform(13))]
}
}
enum Suit: String {
case Spades = "spades"
case Hearts = "hearts"
case Diamonds = "diamonds"
case Clubs = "clubs"
static var randomSuit: Suit {
return [Spades, Hearts, Diamonds, Clubs][Int(arc4random_uniform(4))]
}
}
struct Card: CustomStringConvertible, Equatable {
let number: Number
let suit: Suit
var description: String {
return "\(number.rawValue) of \(suit.rawValue)"
}
static var randomCard: Card {
return Card(number: Number.randomNumber, suit: Suit.randomSuit)
}
static func randomCards(count count: Int) -> [Card] {
guard count > 0 else {
return []
}
guard count <= 52 else {
fatalError("There only are 52 unique cards.")
}
let cards = randomCards(count: count - 1)
while true {
let card = randomCard
if !cards.contains(card) {
return cards + [card]
}
}
}
}
func == (left: Card, right: Card) -> Bool {
return left.number == right.number && left.suit == right.suit
}
let randomCards = Card.randomCards(count: 5)
print(randomCards)
// prints five random cards
如果您有任何其他问题,请告诉我。