快速調用類方法

Question

我創建了一個小程序，其中包含人員姓名，性別和競技場分數，並根據用戶的性別/等級從中打印出一條消息。

在我調用方法之前，一切都很好，我可以看到kemkoi收到了以下值： "kemkoi", "male" and "2500"但是每當我調用我的方法時，它只會顯示"()"

這是代碼：

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

通過執行println( kemkoi.getPlayerStats() )您將打印kemkoi.getPlayerStats()的結果為nil。

嘗試簡單地調用kemkoi.getPlayersStats()

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另請注意，該方法將始終輸出Please choose a gender! 因為您正在執行OR，所以應該改用AND：

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