ios - 无法将类对象分配给 Swift 3 中另一个类对象的字典成员

Question

我有一个 ChessSquare 类，它有自己的属性和方法。我有另一个 ChessBoard 类，它有一个字典类型的属性。

class ChessBoard {
    var squares: [ Int : ChessSquare ] = [:]
    .....

ChessBoard 类的方法之一构建 chessSquare 对象并将其分配给 chessBoard 对象，如下所示。

self.squares[squareName] = chessSquare

在 Swift3 之前，此代码运行良好。升级到 Swift3 后，分配停止工作。

使用断点，我看到 chessSquare 对象是按预期构建的。 squareName 变量具有预期值。 “self”，即 chessBoard 对象，具有适当的初始值。但是在上面提到的代码中，字典分配被跳过而没有任何错误。

这与 Swift3 相关吗？我浏览了但无法找到针对我的案例的具体解决方案。

---

为清楚起见添加更多代码。

class ChessSquare {
    let squareColor: UIColor
    let squareShade: Int
    let squareName: Int
    let squareSize: CGSize
    let minXY: CGPoint
    let maxXY: CGPoint
    let squareOrigin: CGPoint
    var hasPiece = false
    weak var chessPiece: ChessPiece?
    let squareSprite: SKSpriteNode
    let squareType: SquareType
    //more data members

    init(squareColor: UIColor, squareShade: Int, squareName: Int, squareSize: CGSize, minXY: CGPoint, maxXY: CGPoint, squareOrigin: CGPoint, squareSprite: SKSpriteNode, squareType: SquareType) {
        self.squareColor = squareColor
        self.squareShade = squareShade
        self.squareName = squareName
        self.squareSize = squareSize
        self.minXY = minXY
        self.maxXY = maxXY
        self.squareOrigin = squareOrigin
        self.squareSprite = squareSprite
        self.squareType = squareType
    }

    //more methods

}

class ChessBoard {
    var squares: [ Int : ChessSquare ] = [:]
    var byteBoard: [UInt8] = [UInt8](repeating: UInt8(0), count: 66)
    var byteBoardHashVal: Int = 0
    // [ byteBoardHashVal : ((byteBoard, evalnVal), repCnt) ]
    var byteBoards: [Int : (([UInt8],Double?), Int)] = [:]
    var TT: [Int : Double] = [:]
    var TTReuseCount = 0
    var whitePawnlessFileByte: UInt8 = 255
    var blackPawnlessFileByte: UInt8 = 255

    var maxXY: CGPoint = CGPoint()
    var minXY: CGPoint = CGPoint()

    var oldTouchedSquare: ChessSquare?

    init() {
        squares = [:]
        byteBoard = [UInt8](repeating: UInt8(0), count: 66)
        byteBoardHashVal = 0
        byteBoards = [:]
        TT = [:]
        TTReuseCount = 0
        whitePawnlessFileByte = 255
        blackPawnlessFileByte = 255

    }

    func drawFlippedChessBoard(_ view: SKView, scene: GameScene) {
        ....
        ....
        for row in 0..<8 {
            for col in 0..<8 {
                let squareName = row * 8 + col

                ....

                let chessSquare = ChessSquare(squareColor: currentColor, squareShade: squareShade, squareName: squareName, squareSize: squareSize, minXY: minXY, maxXY: maxXY, squareOrigin: squareOrigin, squareSprite: squareSprite, squareType: squareType)

                ....

                self.squares[squareName] = chessSquare

                ....
            }
            ....
        }
        ....
    }

    //more methods

}

Answer 1

我很难理解 squareName 是字符串还是整数。它听起来像一个字符串，我建议将它命名为 chessSquareIndex 或 squareIndex 以防万一它是 Int 并且你想索引国际象棋方 block 。

根据你上面的评论，我做了一个小测试来验证如果 squareName 是 Int 它是否正常工作，这是结果，

class ChessSquare { }

class ChessBoard {
    var squares: [ Int : ChessSquare ] = [:]

    func getChessSquare(at squareIndex: Int) {
        let chessSquare = ChessSquare()
       return squares[squareIndex] = chessSquare
    }
}

这似乎适用于 Swift 3。另请注意，除非您在转义闭包中，否则不需要调用 self。