Question

我创建了2个带相关类型的协议。符合Reader的类型应该能够生成符合Value的类型的实例。

符合Manager的类型的复杂层应该能够生成具体Reader实例，该实例生成特定类型的Value（Value1或Value2）。

通过Manager1的具体实现，我希望始终生成Reader1，然后生成Value1的实例。

有人可以解释原因吗

＆＃34; Reader1无法转换为ManagedReaderType？＆＃34;

当错误的行更改为（现在）返回nil时，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

Answer 1

发生错误是因为ManagerReaderType函数中的read只是符合Reader且其ReaderValueType等于ManagerReaderType的任何类型的通用占位符{1}}。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

如果您返回nil，则可以将其转换为任何可选类型，以便始终有效。

作为替代方案，您可以返回特定类型的类型Reader：

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()
    }
}

这是协议的最佳方法，因为缺少＆＃34; true＆＃34;泛型（以下不支持（还））：

// 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让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?

此实现应解决您的问题，但Readers现在是引用类型，应考虑复制函数。

具有约束关联类型错误的Swift协议＆＃34;类型不可转换＆＃34;

1 个答案: