Sorting Information with Comparable -Swift中实例的排序问题

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Comparable 有两个先决条件：类型要满足Equatable protocol和使用<操作符。

struct Employee: Equatable,Comparable {
        var firstname: String    var lastname: String    var title: String

 //相同实例的先决条件    static func == (lhs: Employee, rhs: Employee) -> Bool {        return lhs.firstname == rhs.firstname && lhs.lastname == rhs.lastname        && lhs.title == rhs.title    }//排序依据    static func < (lhs:Employee, rhs:Employee) -> Bool {                return lhs.lastname < rhs.lastname    }}let employee1 = Employee(firstname: "Si", lastname: "li", title: "PL")let employee2 = Employee(firstname: "Paul", lastname: "maivd", title: "TL")let employee3 = Employee(firstname: "Lulu", lastname: "ain", title: "Sales")let employeeall = [employee1,employee2,employee3]//此处使用闭包的函数 - sorted(by:)let employeesorted = employeeall.sorted(by: <)for employee in employeesorted {        print(employee)}

输出：将lastname 排序后输出结果如下

Employee(firstname: "Lulu", lastname: "ain", title: "Sales")Employee(firstname: "Si", lastname: "li", title: "PL")Employee(firstname: "Paul", lastname: "maivd", title: "TL")

添加Int，可以自由排序（升序或者降序）

struct Employee: Equatable,Comparable {        var firstname: String    var lastname: String    var title: String    var age: Int        static func == (lhs: Employee, rhs: Employee) -> Bool {        return lhs.firstname == rhs.firstname && lhs.lastname == rhs.lastname        && lhs.title == rhs.title    }    static func < (lhs:Employee, rhs:Employee) -> Bool {                return lhs.age < rhs.age    }}let employee1 = Employee(firstname: "Si", lastname: "li", title: "PL", age: 50)let employee2 = Employee(firstname: "Paul", lastname: "maivd", title: "TL", age: 40)let employee3 = Employee(firstname: "Lulu", lastname: "ain", title: "Sales", age: 60)let employeeall = [employee1,employee2,employee3]//此处使用的闭包的函数let employeesorted = employeeall.sorted(by: <)for employee in employeesorted {        print(employee)}

输出结果：

Employee(firstname: "Paul", lastname: "maivd", title: "TL", age: 40)

Employee(firstname: "Si", lastname: "li", title: "PL", age: 50)

Employee(firstname: "Lulu", lastname: "ain", title: "Sales", age: 60)

如果将 sorted(by: <) 改为'>'

输出结果将随之发生变化：

Employee(firstname: "Lulu", lastname: "ain", title: "Sales", age: 60)

Employee(firstname: "Si", lastname: "li", title: "PL", age: 50)

Employee(firstname: "Paul", lastname: "maivd", title: "TL", age: 40)

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