如何使用键值编码访问超类属性？

Question

是否可以使用键值编码来访问对象的超类属性？

我想是这样的：

class Bar: Foo { 
    var shouldOverridePropertyOfFoo: Bool = false 

    var propertyOfFoo: String { 
     if shouldOverrideProperty { 
      return "property of Bar" 
     } else { 
      return value(forKeyPath: "super.propertyOfFoo") as! String 
     } 
    } 
} 

但是，我得到这个运行时错误：

*** Terminating app due to uncaught exception 'NSUnknownKeyException', reason: ' [<MyModule.Bar 0x2da2cf003e00> valueForUndefinedKey:] : this class is not key value coding-compliant for the key super .'

注：我想弄清楚how to override private method and call super in swift?

Answer 1

而不是super.aPropertyOfFoo，你必须使用Foo.aPropertyOfFoo，这可以用#keyPath来代替Strings：

class Foo: NSObject { 
    @objc var aPropertyOfFoo = "property of foo" 
} 

class Bar: Foo { 
    var shouldOverridePropertyOfFoo: Bool = false 

    var propertyOfFoo: String { 
     if shouldOverridePropertyOfFoo { 
      return "property of bar" 
     } else { 
      return value(forKeyPath: #keyPath(Foo.aPropertyOfFoo)) as! String 
     } 
    } 
} 

let bar = Bar() 
print(bar.propertyOfFoo) // "property of foo"