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使用 swift_dynamicCastUnknownClass 将 Swift 泛型转换为 Protocol 失败

转载 作者:搜寻专家 更新时间:2023-10-31 19:35:22 26 4
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以下示例摘自 Apple Swift 引用指南。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,一旦对泛型类型施加约束,它就不再起作用了

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

您的通用函数毫无意义。什么可以解决?什么能满足它?使用相同的基本声明结构做一个更简单的例子:这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()

现在将 g 修改为通用的,与您的函数完全平行:

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。

关于使用 swift_dynamicCastUnknownClass 将 Swift 泛型转换为 Protocol 失败,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28113209/

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