Question

我的课堂上有以下声明：

var peerDevice: [String: Any]? 

private var peerDeviceSettings:[String:Any]? {
    get {
        if let settings = peerDevice?["settings"] as? [String:Any] {
            return settings
        }

        return nil
    }
}

现在的问题是，是否要更改peerDeviceSettings，例如：

 fileprivate func set(_ value: Any?, forKey key: Key) {
    if peerDeviceSettings != nil {
        peerDeviceSettings?[key.rawValue] = value
    } 
 }

我收到错误

 Cannot assign through subscript: 'peerDeviceSettings' is a get-only property

要点是如何重新构建此东西而不创建新的可变变量peerDeviceSettings？

Answer 1

只需将set (setter)添加到属性中并在那里进行处理。

private var peerDeviceSettings:[String:Any]? {
    get {
        if let settings = peerDevice?["settings"] as? [String:Any] {
            return settings
        }

        return nil
    }
    set {

    }
}

Answer 2

您的代码似乎很好，因为您正在创建到嵌套settings字典的链接（听起来合理）

只需添加二传手：

private var peerDeviceSettings: [String: Any]? {
    get {
        if let settings = peerDevice?["settings"] as? [String:Any] {
            return settings
        }

        return nil
    }
    set { 
      if peerDevice == nil {
        peerDevice = [:] // if it's desired behavior
      }
      peerDevice?["settings"] = newValue 
    }

}

我还建议您摆脱peerDevice中的可选性，只需将其初始化为空值（如果它符合您的逻辑）

Answer 3

您看到此错误，因为您仅声明了属性的getter，还添加了setter

    private var peerDeviceSettings:[String:Any]? {
    get {
        if let settings = peerDevice?["settings"] as? [String:Any] {
            return settings
        }

        return nil
    }
    set {
        peerDevice?["settings"] = newValue
    }
}

更新：显示在特定情况下必须如何显示

Answer 4

您需要二传手

private var peerDeviceSettings:[String:Any]? {
    get {
        return  peerDevice?["settings"] as? [String:Any]
    }

    set {
        peerDevice?["settings"] = newValue
    }
}

Swift无法通过下标分配：“ peerDeviceSettings”是仅获取属性”

4 个答案: