iOS Swift4 如何保护接口(interface)中定义的计算属性不被赋值？

Question

我正在为我继承的遗留代码库中的类调试一些编写不当的单元测试。我看到开发人员没有正确测试计算属性。我想了解如何强制我的 userEnabledFeature 在 Fake 类中只读。

为什么使用计算属性实现协议(protocol)的类可以覆盖计算属性并使其可写？

public protocol FeatureManager {
    var userEnabledFeature : Bool { get }
}

public class FakeFeatureManagerForTesting: FeatureManager {
    public var userEnabledFeature = false //is this legal? Why is compiler not complaining?

    public func updateUserEnabledFeature(enabled: Bool){
        //this should not be possible - how do I prevent overwriting computed property?
        userEnabledFeature = enabled
    }
}

ActualFeatureManagerClass {
    public var userEnabledFeature: Bool {
        if featureManager.cachedFeatures.filter { $0.enabled == true}
        {
            //do more checks, return true or false
        }
        return false //default
    }
}

Answer 1

public class FakeFeatureManagerForTesting: FeatureManager {
  public var userEnabledFeature = false //is this legal? Why is compiler not complaining?
}

协议(protocol)是说符合它的成员必须有一个Bool属性userEnabledFeature可以读取，并不是说它是只读属性。但是，如果您有一个以协议(protocol)为类型的变量，即使变量的值是实现也具有可写性，它也不允许您分配属性。

协议(protocol)没有定义它是如何实现的(如果它的计算属性、存储属性等)，但它的签名是什么以及它是应该只读还是必须 也有写入权限。如果您将其标记为只读，该实现仍然可以使写可写，因为在这种情况下协议(protocol)并不真正关心写访问。

您可以像这样实现计算属性:

public class FakeFeatureManagerForTesting: FeatureManager {
  public var userEnabledFeature: Bool { 
    // do some work here and return value or fallback to false
    return false
  }
}

另一方面，您也可以只有私有(private)可设置属性，只能从文件或类/结构更改:

public class FakeFeatureManagerForTesting: FeatureManager {
  public private(set) var userEnabledFeature = false

  public func updateUserEnabledFeature(enabled: Bool) {
    userEnabledFeature = enabled // will work just fine
  }
}

let manager = FakeFeatureManagerForTesting()
manager.updateUserEnabledFeature(enabled: true) // will work
manager.userEnabledFeature = true // won't compile, because modifying is allowed privately only