swift2 Swift 3中的动态类型

bq3bfh9z 于 2022-11-06 发布在 Swift

关注(0)|答案(4)|浏览(233)

我已经将我的swift版本从2.3迁移到3，它自动转换了一些代码，下面是我遇到崩溃的情况，我尝试了一些选项，但都是徒劳的，
swift 2.3：* 工作正常 *

public func huntSuperviewWithClassName(className: String) -> UIView?
{
    var foundView: UIView? = nil

    var currentVeiw:UIView? = self

    while currentVeiw?.superview != nil{
        if let classString = String.fromCString(class_getName(currentVeiw?.dynamicType)){

            if let classNameWithoutPackage = classString.componentsSeparatedByString(".").last{
                print(classNameWithoutPackage)
                if classNameWithoutPackage == className{
                    foundView = currentVeiw
                    break
                }
            }
        }
        currentVeiw = currentVeiw?.superview
    }

    return foundView
}

}
雨燕3：不好

if let classString = String(validatingUTF8: class_getName(type(of:currentVeiw) as! AnyClass)) {

也试过这一行：

if let classString = String(describing: class_getName(type(of: currentVeiw) as! AnyClass)){

但它不起作用。
请指导我如何根据swift3：

if let classString = String.fromCString(class_getName(currentVeiw?.dynamicType)){

swift2

来源：https://stackoverflow.com/questions/45790180/dynamic-type-in-swift-3

4条答案

按热度按时间

pgccezyw1#

编译器告诉你不能使用if let，因为它是完全不必要的。你没有任何可选择的解包。if let专门用于解包可选择的。

public func huntSuperviewWithClassName(className: String) -> UIView?
{
    var foundView: UIView? = nil

    var currentVeiw:UIView? = self

    while currentVeiw?.superview != nil{

            let classString = NSStringFromClass((currentVeiw?.classForCoder)!)

            if let classNameWithoutPackage = classString.components(separatedBy:".").last{
                print(classNameWithoutPackage)
                if classNameWithoutPackage == className{
                    foundView = currentVeiw
                    break
                }
            }
        }
        currentVeiw = currentVeiw?.superview
    }

    return foundView
}

工作正常！

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bttbmeg02#

if let classString = String(describing: currentVeiw.self) 
{
}

赞(0）回复(0）举报 2022-11-06

gc0ot86w3#

只需执行以下操作：

let classString = String(describing: type(of: currentVeiw!))

赞(0）回复(0）举报 2022-11-06

voj3qocg4#

请尝试以下操作：

public func huntSuperviewWithClassName(className: String) -> UIView?
{
    var foundView: UIView? = nil
    var currentVeiw:UIView? = self
    while currentVeiw?.superview != nil{
        let classString = String(describing: type(of: currentVeiw?.classForCoder))
        if let classNameWithoutPackage = classString.components(separatedBy:".").first {
            print(classNameWithoutPackage)
            if classNameWithoutPackage == className {
                foundView = currentVeiw
                break
            }
        }
        currentVeiw = currentVeiw?.superview
    }
    return foundView
}

赞(0）回复(0）举报 2022-11-06

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swift2 Swift 3中的动态类型

4条答案

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