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

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
4000
", 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)