快速错误: Could not find an overload for '&&' that accepts the supplied arguments

Question

我一直在编写一些 Swift 教程。我遇到了一个 TicTacToe 教程，我正在尝试使用 Xcode 6 Beta 6 进行编码。当我检查字典中的值时，出现以下错误:无法找到接受提供的参数的“&&”的重载。这是我的代码。

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!"
 }

Answer 1

我正在做同样的教程。在我看来，它必须被分解为子表达式有点奇怪，但它确实做到了，正如我在这里读到的: http://swiftlang.eu/community/34-xcode-6-beta-6-swift-can-t-handle-long-expressions/0

可能是Xcode的问题。

这是重写的函数:

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
        }
    }
}