Swift文档指出,类、结构体和枚举都可以遵循协议,并且我可以使它们都符合要求。但是我无法让枚举的行为与类和结构体的示例完全相同:
protocol ExampleProtocol {
var simpleDescription: String { get set }
mutating func adjust()
}
class SimpleClass: ExampleProtocol {
var simpleDescription: String = "A very simple class."
var anotherProperty: Int = 69105
func adjust() {
simpleDescription += " Now 100% adjusted."
}
}
var a = SimpleClass()
a.adjust()
let aDescription = a.simpleDescription
struct SimpleStructure: ExampleProtocol {
var simpleDescription: String = "A simple structure"
mutating func adjust() {
simpleDescription += " (adjusted)"
}
}
var b = SimpleStructure()
b.adjust()
let bDescription = b.simpleDescription
enum SimpleEnum: ExampleProtocol {
case Base
var simpleDescription: String {
get {
return "A Simple Enum"
}
set {
newValue
}
}
mutating func adjust() {
self.simpleDescription += ", adjusted"
}
}
var c = SimpleEnum.Base
c.adjust()
let cDescription = c.simpleDescription
我还没有想出如何在调用adjust()
后更改simpleDescription
的方法。我的示例显然无法做到这一点,因为getter具有硬编码的值,但是我如何在仍符合ExampleProtocol
的情况下设置simpleDescription
的值呢?
ExampleProtocol
中的adjust
函数返回Void
,所以与使用mutating func adjust()
相同。如果您想让adjust
具有返回类型,可以将协议更改为:https://gist.github.com/anjerodesu/e1bf640576a3b6fa415f - Angelocase .Base:
。 - John Doe