ios - 如何编码 [字符 : Int] property using NSCoder in Swift 3?

Question

import UIKit

class Foo: NSObject, NSCoding {
    var cx: [Character : Int]

    init(cx: [Character : Int]) {
        self.cx = cx
    }

    // MARK: - <NSCoding>

    required convenience init(coder aDecoder: NSCoder) {
        let cx = aDecoder.decodeObject(forKey: "cxKey") as! [Character : Int]
        self.init(cx: cx)
    }

    func encode(with aCoder: NSCoder) {
        aCoder.encode(cx, forKey: "cxKey")
    }
}

调用:

class ViewController: UIViewController {

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view, typically from a nib.

        var foo = Foo(cx: ["C": 5, "X": 6])

        let encodedData = NSKeyedArchiver.archivedData(withRootObject: foo)
        print("encodedData: \(encodedData))")

        if let foo1 = NSKeyedUnarchiver.unarchiveObject(with: encodedData) as? Foo {
            print("cx = ", foo1.cx)
        } else{
            print("There is an issue")
        }
    }
}

Xcode 抛出错误:*** 由于未捕获的异常“NSInvalidArgumentException”而终止应用程序，原因:“-[_SwiftValue encodeWithCoder:]: 无法识别的选择器发送到实例

Answer 1

原因

这是因为 cx 中的 Character 类型的键将被装箱为 _SwiftValue 对象，该对象将被发送 encodeWithCoder: 导致无法识别的选择器异常。

请参阅 SwiftValue.h 顶部的评论:

This implements the Objective-C class that is used to carry Swift values that have been bridged to Objective-C objects without special handling. The class is opaque to user code, but is NSObject- and NSCopying- conforming and is understood by the Swift runtime for dynamic casting back to the contained type.

解决方案

如果您可以将 cx 的类型更改为 [String : Int]，一切都会开箱即用(无双关语意)。

否则，您必须将 Foo.encode(with:) 中的 cx 转换为可以编码的内容(如 [String : Int]，例如)，反之亦然。

参见 How do I encode Character using NSCoder in swift?和 How do I encode enum using NSCoder in swift?对于一些代码。