以下示例取自Apple
Swift参考指南.我只添加了getHasAreaInstances()和getGenericHasAreaInstances()
import UIKit
@objc protocol HasArea {
var area: Double { get }
}
@objc protocol HasExtendedArea: HasArea {
var extendedArea: Double { get }
}
class Circle: HasArea {
let pi = 3.1415927
var radius: Double
var area: Double { return pi * radius * radius }
init(radius: Double) { self.radius = radius }
}
class Country: HasArea {
var area: Double
init(area: Double) { self.area = area }
}
class Continent: HasExtendedArea {
var area: Double { return 300 }
var extendedArea: Double { return 3000 }
}
let objects: [HasArea] = [
Circle(radius: 2.0),
Country(area: 243_610),
Continent()
]
for object in objects {
if let objectWithArea = object as? HasExtendedArea {
println("Extended Area is \(objectWithArea.area)")
} else {
println("Area is not extended")
}
}
// Extended Area is 300.0
// Area is not extended
// Area is not extended
下面的方法返回正确的数组:
func getHasExtendedAreaInstances() -> [HasExtendedArea] {
var haveArea: [HasExtendedArea] = []
for object in objects {
if let objectWithArea = object as? HasExtendedArea {
haveArea.append(objectWithArea)
}
}
return haveArea
}
let areas = getHasExtendedAreaInstances()
//[Continent]
下面的方法返回正确的数组:
func getGenericHasExtendedAreaInstances<T>() -> [T] {
var haveArea: [T] = []
for object in objects {
if let objectWithArea = object as? T {
haveArea.append(objectWithArea)
}
}
return haveArea
}
let areasGeneric: [HasExtendedArea] = getGenericHasExtendedAreaInstances()
//[Continent]
但是,只要对泛型类型施加约束,它就不再起作用
func getGenericConstraintHasExtendedAreaInstances<T: HasArea>() -> [T] {
var haveArea: [T] = []
for object in objects {
if let objectWithArea = object as? T {
// the line above fails with swift_dynamicCastUnknownClass
haveArea.append(objectWithArea)
}
}
return haveArea
}
let areasGenericConstraint: [HasExtendedArea] = getGenericConstraintHasExtendedAreaInstances()
最佳答案 你的通用功能毫无意义.什么能解决它?什么会满足它?使用相同的基本声明结构制作一个更简单的示例:它是一个不可能的函数.例如,从这个非泛型函数开始:
class Thing : Printable {
var description : String {return "thing"}
}
func g() -> [Thing] {
return [Thing()]
}
let result : [Thing] = g()
现在修改g是通用的,与你的函数完全平行:
class Thing : Printable {
var description : String {return "thing"}
}
func g<T:Printable>() -> [T] {
return [Thing()]
}
let result : [Thing] = g()
它不编译 – 因为它没有意义.