[英]Swift protocol with constrained associated type error “Type is not convertible”
我创建了2个具有相关类型的协议。 符合Reader
的类型应该能够生成符合Value
的类型的实例。
复杂层来自符合Manager
的类型,应该能够生成一个具体的Reader
实例,它生成一个特定类型的Value
( Value1
或Value2
)。
通过我对Manager1
具体实现,我希望它始终生成Reader1
,而Reader1
又生成Value1
实例。
有人可以解释原因
“Reader1无法转换为ManagedReaderType?”
当错误的行改为(现在)返回nil
,所有编译都很好,但现在我无法实例化Reader1
或Reader2
。
可以将以下内容粘贴到Playground中以查看错误:
import Foundation
protocol Value {
var value: Int { get }
}
protocol Reader {
typealias ReaderValueType: Value
func value() -> ReaderValueType
}
protocol Manager {
typealias ManagerValueType: Value
func read<ManagerReaderType: Reader where ManagerReaderType.ReaderValueType == ManagerValueType>() -> ManagerReaderType?
}
struct Value1: Value {
let value: Int = 1
}
struct Value2: Value {
let value: Int = 2
}
struct Reader1: Reader {
func value() -> Value1 {
return Value1()
}
}
struct Reader2: Reader {
func value() -> Value2 {
return Value2()
}
}
class Manager1: Manager {
typealias ManagerValueType = Value1
let v = ManagerValueType()
func read<ManagerReaderType: Reader where ManagerReaderType.ReaderValueType == ManagerValueType>() -> ManagerReaderType? {
return Reader1()// Error: "Reader1 is not convertible to ManagedReaderType?" Try swapping to return nil which does compile.
}
}
let manager = Manager1()
let v = manager.v.value
let a: Reader1? = manager.read()
a.dynamicType
发生此错误是因为read
函数中的ManagerReaderType
只是符合Reader
且其ReaderValueType
等于ManagerReaderType
任何类型的通用占位符。 因此, ManagerReaderType
的实际类型不是由函数本身决定的,而是被赋值的变量的类型声明了类型:
let manager = Manager1()
let reader1: Reader1? = manager.read() // ManagerReaderType is of type Reader1
let reader2: Reader2? = manager.read() // ManagerReaderType is of type Reader2
如果你返回nil
它可以转换为任何可选类型,所以它总是有效。
作为替代方案,您可以返回特定类型的Reader
类型:
protocol Manager {
// this is similar to the Generator of a SequenceType which has the Element type
// but it constraints the ManagerReaderType to one specific Reader
typealias ManagerReaderType: Reader
func read() -> ManagerReaderType?
}
class Manager1: Manager {
func read() -> Reader1? {
return Reader1()
}
}
由于缺少“真正的”泛型,这是使用协议的最佳方法(不支持以下内容):
// this would perfectly match your requirements
protocol Reader<T: Value> {
fun value() -> T
}
protocol Manager<T: Value> {
func read() -> Reader<T>?
}
class Manager1: Manager<Value1> {
func read() -> Reader<Value1>? {
return Reader1()
}
}
所以最好的解决方法是使Reader
成为泛型类, Reader1
和Reader2
子类是它的特定泛型类型:
class Reader<T: Value> {
func value() -> T {
// or provide a dummy value
fatalError("implement me")
}
}
// a small change in the function signature
protocol Manager {
typealias ManagerValueType: Value
func read() -> Reader<ManagerValueType>?
}
class Reader1: Reader<Value1> {
override func value() -> Value1 {
return Value1()
}
}
class Reader2: Reader<Value2> {
override func value() -> Value2 {
return Value2()
}
}
class Manager1: Manager {
typealias ManagerValueType = Value1
func read() -> Reader<ManagerValueType>? {
return Reader1()
}
}
let manager = Manager1()
// you have to cast it, otherwise it is of type Reader<Value1>
let a: Reader1? = manager.read() as! Reader1?
此实现应该可以解决您的问题,但Readers
现在是引用类型,应该考虑复制函数。
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