Swift错误：找不到'&&'；它接受提供的参数

我一直在为斯威夫特编写一些教程。我遇到了一个TictaToe教程，我正试图使用Xcode 6 Beta 6编写该教程。在检查字典中的值时，出现以下错误：找不到接受所提供参数的“&&”的重载。这是我的密码

var plays = [Int:Int]()

var whoWon = ["I":0,"you":1]
for (key,value) in whoWon {
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[6] == value && plays[3] == value && plays[0] == value) || 
    (plays[7] == value && plays[4] == value && plays[1] == value) || 
    (plays[8] == value && plays[5] == value && plays[2] == value) || 
    (plays[6] == value && plays[4] == value && plays[2] == value) ||  // error appears on this line
    (plays[8] == value && plays[4] == value && plays[0] == value))  
 {
    userMessage.hidden = false
    userMessage.text = "Looks like \(key) won!"
 }

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如果您在报表导航器中查看完整的编译器输出，那么您将看到 信息

注：表达式过于复杂，无法在合理时间内解决； 把表达式分解成不同的子表达式

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它告诉你如何解决这个问题。

我也在做同样的教程。 在我看来，有点奇怪，它必须被分解成子表达式，但它做到了，正如我在这里读到的：

这可能是由于Xcode的问题

下面是重写的函数：

func checkForWin() {
    var whoWon = ["I": 0, "You": 1]
    for (key, value) in whoWon {
        var wonA = (plays[1] == value && plays[2] == value && plays[3] == value)
        var wonB = (plays[4] == value && plays[5] == value && plays[6] == value)
        var wonC = (plays[7] == value && plays[8] == value && plays[9] == value)
        var wonD = (plays[1] == value && plays[4] == value && plays[7] == value)
        var wonE = (plays[2] == value && plays[5] == value && plays[8] == value)
        var wonF = (plays[3] == value && plays[6] == value && plays[9] == value)
        var wonG = (plays[1] == value && plays[5] == value && plays[9] == value)
        var wonH = (plays[3] == value && plays[5] == value && plays[7] == value)

        if(wonA || wonB || wonC || wonD || wonE || wonF || wonG || wonH) {
                userMessage.hidden = false
                userMessage.text = "Looks like \(key) won!"
                resetBtn.hidden = false
                done = true
        }
    }
}

使用括号“将表达式分解为不同的子表达式”。 这在Xcode 6.4中起作用

 if (((plays[1] == value) && (plays[2] == value) && (plays[3] == value)) ||
     ((plays[4] == value) && (plays[5] == value) && (plays[6] == value)) ||
     ((plays[7] == value) && (plays[8] == value) && (plays[9] == value)) ||
     ((plays[1] == value) && (plays[4] == value) && (plays[7] == value)) ||
     ((plays[2] == value) && (plays[5] == value) && (plays[8] == value)) ||
     ((plays[3] == value) && (plays[6] == value) && (plays[9] == value)) ||
     ((plays[1] == value) && (plays[5] == value) && (plays[9] == value)) ||
     ((plays[3] == value) && (plays[5] == value) && (plays[7] == value))) {...}

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谢谢我忘记了“问题导航器”选项卡。嘿@Martin R我正在使用同一个教程，我是一个初学者，你如何将它分解为子表达式？如何将某些内容分解为子表达式？为什么

var

而不是

let

？（即，won*可以是常量，对吗？…）