Swift：更新Xcode:&x27时出错；找不到'&&'；接受提供的参数'；

我刚刚更新到Xcode 6的非测试版（最终），从测试版5开始，出现了一些我以前没有遇到的错误，其中一个错误是“找不到接受所提供参数的“&&”的重载”

我在这里学习一个教程，从另一个问题中，我知道这个错误是因为“表达式太复杂，无法在合理的时间内解决；请考虑将表达式分解为不同的子表达式。” 我是初学者；如何将表达式分解为子表达式

我的代码：

func checkForWin(){
        //first row across
        var youWin = 1
        var theyWin = 0
        var whoWon = ["Lost":0,"Won":1]
        for (key,value) in whoWon {
            if ((plays[6] == value && plays[7] == value && plays[8] == value) || //across the bottom
            (plays[3] == value && plays[4] == value && plays[5] == value) || //across the middle
            (plays[0] == value && plays[1] == value && plays[2] == value) || //across the top
            (plays[6] == value && plays[3] == value && plays[0] == value) || //down the left side
            (plays[7] == value && plays[4] == value && plays[1] == value) || //down the middle
            (plays[8] == value && plays[5] == value && plays[2] == value) || //down the right side
            (plays[6] == value && plays[4] == value && plays[2] == value) || //diagonal
                (plays[8] == value && plays[4] == value && plays[0] == value)){//diagonal
                    userMessage.hidden = false
                    youLabel.hidden = false
                    userMessage.text = "\(key)!"
                    done = true;
            }
        }
    }

---

要分解它，只需添加更多（）即可。像这样：

 if (((plays[6] == value) && (plays[7] == value) && (plays[8] == value)) ||
 ((plays[3] == value) && (plays[4] == value) && (plays[5] == value)) ||
 ((plays[0] == value) && (plays[1] == value) && (plays[2] == value)) ||

 ((plays[8] == value) && (plays[4] == value) && (plays[0] == value)))

var plays=[Int:Int]（）