Swift AnyObject的下标,它来自哪里？

Question

在Swift中,它是如何AnyObject支持下标的,即使对于不可下载的类型也是如此？例:

let numbers: AnyObject = [11, 22, 33]
numbers[0]       // returns 11

let prices: AnyObject = ["Bread": 3.49, "Pencil": 0.5]
prices["Bread"]  // returns 3.49

let number: AnyObject = 5
number[0]        // return nil

let number: AnyObject = Int(5)
number[0]        // return nil

然而,如果我number被声明为Int那么这是一个语法错误:

let number: Int = 5
number[0]        // won't compile

有趣的是,Any没有下标支持.

Answer 1

这只适用于导入Foundation,因为Swift在这种情况下正在做一些桥接Objective-C类型 - NSArray在这种情况下是类似对象.

import Foundation

let numbers: AnyObject = [11, 22, 33] as AnyObject
type(of: numbers) //_SwiftDeferredNSArray.Type

如果不导入Foundation,则甚至不允许进行赋值(因为Swift数组是结构而不是对象).

let numbers: AnyObject = [11, 22, 33] // error: contextual type 'AnyObject' cannot be used with array literal

你可以投Any,但:

let numbers: Any = [11, 22, 33]    
type(of: numbers) // Array<Int>.Type

为什么这样import Foundation做？这在AnyObject类型描述中记录:

/// When used as a concrete type, all known `@objc` methods and  
/// properties are available, as implicitly-unwrapped-optional methods  
/// and properties respectively, on each instance of `AnyObject`.  For  
/// example:  
///  
///     class C {  
///       @objc func getCValue() -> Int { return 42 }  
///     }  
///  
///     // If x has a method @objc getValue()->Int, call it and  
///     // return the result.  Otherwise, return nil.  

这意味着你甚至可以在数组上调用不一定存在NSArray但在Objective-C世界中存在的方法,例如:

numbers.lowercaseString       // nil

并且Swift将优雅地返回一个nil值而不是抛出一个讨厌的object does not recognises selector异常,就像在Objective-C中会发生的那样.如果这是好还是坏,还是要辩论:)

更新 以上似乎只适用于属性和类似属性的方法,如果您尝试使用Objective-C方法,那么您将遇到无法识别的选择器问题:

import Foundation

@objc class TestClass: NSObject {
    @objc var someProperty: Int = 20
    @objc func someMethod() {}
}

let numbers: AnyObject = [11, 22, 33] as AnyObject
numbers.lowercaseString                // nil
numbers.someMethod                     // nil
numbers.someMethod()                   // unrecognized selector
numbers.stringByAppendingString("abc") // unrecognized selector

Answer 2

在为类型对象赋值时,它与类型的桥接有关AnyObject:

let numbers: AnyObject = [11, 22, 33]
print(numbers.dynamicType) // _SwiftDeferredNSArray.Type

let prices: AnyObject = ["Bread": 3.49, "Pencil": 0.5]
print(prices.dynamicType) // _NativeDictionaryStorageOwner<String, Double>.Type

let number: AnyObject = 5
print(number.dynamicType) // __NSCFNumber.Type

let anotherNumber: Int = 5
print(anotherNumber.dynamicType)  // Int.Type

在幕后,_SwiftDeferredNSArray.Type,_NativeDictionaryStorageOwner<String, Double>.Type,并且__NSCFNumber.Type必须同时支持标Int.Type没有.

这假设您已导入Foundation.有关纯Swift类型的解释,请参阅Cristik的答案.