Ios “Any”不能转换为“UIView”

创建一个函数，使网格中的图像随机化。以下是导致问题的swift代码的一部分。感谢您的帮助

    class ViewController: UIViewController {
    var allImgViews = [Any]()
    var allCenters = [Any]()

    override func viewDidLoad() {
    //code
    random()
    }

    func random() {
    var centesCopy: [Any] = allCenters
    var randLocInt: Int
    var randLoc: CGPoint
    for any: UIView in allImgViews {
        randLocInt = arc4random() % centersCopy.count
        randLoc = allCenters[randLocInt].cgPoint()
        any.center = randLoc
    }
}
   }

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您在代码中选择“any”有什么具体原因吗

因为就我在代码中所见，AllimgView似乎是UIImageView的数组，allCenters是CGPoints的数组

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尽可能使用特定的数据类型。仅当您计划使用任意子类的不同类型的对象时才使用Any。

您要查找的代码是

for view in allImgViews as! [UIView] {
        // add your logic here
        // 'view' here is a UIView object
    }

建议：如果您已经知道数组将只包含UIView对象，那么最好像下面那样声明它

var allImgViews = [UIView]()

for view in allImgViews {
    // here the compiler already knows 'view' is a UIView object
    // no need for type casting. 
    //add your logic. 
}

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如果确实不想将allCenter声明为CGPoint，将allImgViews声明为UIImageView，可以尝试以下操作

func random() {

        var centesCopy: [Any] = allCenters
        var randLocInt: Int
        var randLoc: CGPoint
        for index in stride(from: 0, to: allImgViews.count, by: 1)  {
            randLocInt = Int (arc4random()) % centesCopy.count
            randLoc = self.allCenters[randLocInt] as! CGPoint
            var any = UIView()
            any.center = randLoc
            allImgViews[index] = any
        }

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你为什么用[任何]？如果数组包含UIView，则将其声明为[UIView]谢谢您的帮助谢谢您的帮助很高兴它帮助了您。