在Swift中符合Comparable的泛型类 [英] Generic class that conforms to Comparable in Swift

问题描述

我试图创建一个符合Comparable协议的简单通用节点类，以便我可以在不访问密钥的情况下轻松比较节点。当我尝试写入<和==功能，但是，编译器似乎不喜欢它。 <和==函数在定义Node参数时需要一个类型。这在Java中很简单，您定义了平等和<内部向班级。 Swift在全球范围内寻求它。任何想法？

I'm attempting to create a simple generic node class that conforms to the Comparable protocol so that I can easily compare nodes without accessing their key. When I attempt to write the < and == functions, however, the compiler doesn't seem to like it. The < and == functions expect a type when defining the Node parameters. This was simple in Java where you defined equality and < internally to the class. Swift asks for it globally. Any thoughts ?

示例：

Example:

func < (lhs:Node<E:Comparable>, rhs:Node<E:Comparable>) -> Bool {
    return lhs.key < rhs.key
}


func == (lhs:Node<E:Comparable>, rhs:Node<E:Comparable>) -> Bool {
    return lhs.key == rhs.key
}


class Node<D:Comparable>: Comparable {

    var key: D!
    var next:Node?
    var prev:Node?

    init( key:D ) {

        self.key = key
    }
}

推荐答案

你很近！ Node类已经指定对于 Node ， D 必须符合 Comparable 。因此， == 和 <<中的节点< E：Comparable> 是多余的。相反，您希望限制可以调用操作符的类型：



You're close! The Node class already specifies that for Node<D>, D must conform to Comparable. Therefore, Node<E: Comparable> in the decl for == and < is redundant. Instead, you want to restrict the types that the operators can be invoked upon:

func < <E: Comparable>(lhs: Node<E>, rhs: Node<E>) -> Bool {
    return lhs.key < rhs.key
}


func == <E: Comparable>(lhs: Node<E>, rhs: Node<E>) -> Bool {
    return lhs.key == rhs.key
}

class Node<D: Comparable>: Comparable {

    var key: D!
    var next: Node?
    var prev: Node?

    init(key: D) {
        self.key = key
    }
}

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