具有约束关联类型错误的Swift协议“类型不可转换”

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,但读者现在是引用类型,应该考虑复制函数.