快速调用类方法

Question

我创建了一个小程序，其中包含人员姓名，性别和竞技场分数，并根据用户的性别/等级从中打印出一条消息。

在我调用方法之前，一切都很好，我可以看到kemkoi收到了以下值： "kemkoi", "male" and "2500"但是每当我调用我的方法时，它只会显示"()"

这是代码：

class Player {
    var name : String
    var gender : String
    var arenaPoints : Int

    // method
    func getPlayerStats() {

        if(gender != "male" || gender != "female") {
            println("Please choose a gender!")
        } else if(arenaPoints < 2000) {
            println("Seems like you're too unexperienced \(arenaPoints) rating is not enough, come back when you have over 2.2k!")
        } else {
            println("Hello, \(name). You've chosen to play as a \(gender) and your arena points are equal to: \(arenaPoints)")
        }
    }

    // initializer
    init(name : String, gender : String, arenaPoints : Int) {
        self.name = name
        self.gender = gender
        self.arenaPoints = arenaPoints
    }

}

var kemkoi = Player(name: "kemkoi", gender: "male", arenaPoints: 2500)

println( kemkoi.getPlayerStats() )

Answer 1

通过执行println( kemkoi.getPlayerStats() )您将打印kemkoi.getPlayerStats()的结果为nil。

尝试简单地调用kemkoi.getPlayersStats()

---

另请注意，该方法将始终输出Please choose a gender! 因为您正在执行OR，所以应该改用AND：

if(gender != "male" && gender != "female")