xcode 如何使用 NSCoding 将 Int 编码为可选
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How to encode Int as an optional using NSCoding
提问by Eatton
I am trying to declare two properties as optionals in a custom class - a String and an Int.
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'm doing this in MyClass:
我在 MyClass 中这样做：
var myString: String?
var myInt: Int?
I can decode them ok as follows:
我可以按如下方式解码它们：
required init?(coder aDecoder: NSCoder) {
myString = aDecoder.decodeObjectForKey("MyString") as? String
myInt = aDecoder.decodeIntegerForKey("MyInt")
}
But encoding them gives an error on the Int line:
但是对它们进行编码会在 Int 行上出现错误：
func encodeWithCoder(aCoder: NSCoder) {
aCoder.encodeInteger(myInt, forKey: "MyInt")
aCoder.encodeObject(myString, forKey: "MyString")
}
The error only disappears when XCode prompts me to unwrap the Int as follows:
只有当 XCode 提示我解开 Int 时，错误才会消失，如下所示：
aCoder.encodeInteger(myInt!, forKey: "MyInt")
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?
但这显然会导致崩溃。所以我的问题是，如何让 Int 像 String 一样被视为可选项？我错过了什么？
回答by Sulthan
If it can be optional, you will have to use encodeObjectfor it, too.
如果它可以是可选的，那么您也必须使用encodeObject它。
You are using an Objective-C framework and Objective-C allows nilonly for objects (class/reference types). An integer cannot be nilin Objective-C.
您使用的是 Objective-C 框架，而 Objective-Cnil仅允许对象（类/引用类型）。整数不能nil在 Objective-C 中。
However, if you use encodeObject, Swift will automatically convert your Intto NSNumber, which can be nil.
但是，如果您使用encodeObject，Swift 会自动将您转换Int为NSNumber，也可以是nil.
Another option is to skip the value entirely:
另一种选择是完全跳过该值：
if let myIntValue = myInt {
aCoder.encodeInteger(myIntValue, forKey: "MyInt")
}
and use containsValueForKey(_:)when decoding.
并containsValueForKey(_:)在解码时使用。
