如何使用NSCoding将Int编码为可选 [英] How to encode Int as an optional using NSCoding

问题描述

我试图在自定义类中将两个属性声明为可选属性-String和Int.

I am trying to declare two properties as optionals in a custom class - a String and an Int.

我正在MyClass中这样做:

I'm doing this in MyClass:

var myString: String?
var myInt: Int?

我可以按如下所示对它们进行解码:

I can decode them ok as follows:

required init?(coder aDecoder: NSCoder) {
  myString = aDecoder.decodeObjectForKey("MyString") as? String
  myInt = aDecoder.decodeIntegerForKey("MyInt")
}

但是对它们进行编码会在Int行上产生错误:

But encoding them gives an error on the Int line:

func encodeWithCoder(aCoder: NSCoder) {
  aCoder.encodeInteger(myInt, forKey: "MyInt")
  aCoder.encodeObject(myString, forKey: "MyString")
}

仅当XCode提示我打开Int时，错误才会消失:

The error only disappears when XCode prompts me to unwrap the Int as follows:

  aCoder.encodeInteger(myInt!, forKey: "MyInt")

但这显然会导致崩溃.所以我的问题是，如何使Int像String一样被视为可选内容?我想念什么?

But that obviously results in a crash. So my question is, how can I get the Int to be treated as an optional like the String is? What am I missing?

推荐答案

如果可以选择，也必须使用encodeObject.

If it can be optional, you will have to use encodeObject for it, too.

您正在使用Objective-C框架，而Objective-C仅允许nil用于对象(类/引用类型).在Objective-C中，整数不能为nil.

You are using an Objective-C framework and Objective-C allows nil only for objects (class/reference types). An integer cannot be nil in Objective-C.

但是，如果您使用encodeObject，Swift将自动将您的Int转换为NSNumber，可以是nil.

However, if you use encodeObject, Swift will automatically convert your Int to NSNumber, which can be nil.

另一种选择是完全跳过该值:

Another option is to skip the value entirely:

if let myIntValue = myInt {
    aCoder.encodeInteger(myIntValue, forKey: "MyInt")
}

，并在解码时使用containsValueForKey(_:).

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