探讨:如何使用委托,匿名方法对集合进行万能排序

程序员文章站 2024-02-13 11:08:22

下面demo中我使用了2种排序方式1.让employee继承icomparable 接口，实现compareto方法排序2.定义一个方法使用comparison

下面demo中我使用了2种排序方式
1.让employee继承icomparable 接口，实现compareto方法排序
2.定义一个方法使用comparison<t>委托，排序的操作交给匿名方法
看完下面的代码，你知道使用comparison<t>委托的好处吗？

复制代码代码如下:

    class employee:icomparable
    {
        public string name { get; set; }
        public int age { get; set; }
        public static list<employee> getemployees()
        {
            return new list<employee>()
            {
                new employee(){name ="guohu",age =25},
                new employee(){name ="leihu",age =23},
                new employee(){name ="junwenli",age =24},
                new employee(){name ="jinhaoliu",age =25},
                new employee(){name ="chengfang",age =24}
            };
        }
        public int compareto(object obj)
        {
            employee employee = obj as employee;
            if (employee != null)
            {
                return name.compareto(employee.name);
            }
            else
            {
                throw new argumentexception("obj is not employee");
            }
        }
    }
    class test
    {
        static void sortandshowfiles(string title, comparison<employee> employeeinfo)
        {
            list<employee> employee = employee.getemployees();
            employee.sort(employeeinfo);
            console.writeline(title);
            foreach(employee e in employee)
            {
                console.writeline("name:{0},age:{1}", e.name, e.age);
            }
        }
        static void main()
        {
            list<employee> employeeinfo = employee.getemployees();
            //using icomparable sort 
            employeeinfo.sort();
            employeeinfo.foreach(e => console.writeline("name:{0},age{1}/t", e.name, e.age));
            sortandshowfiles("sort by name", delegate(employee e1, employee e2) { return e1.name.compareto(e2.name); });
            sortandshowfiles("sort by age", delegate(employee e1, employee e2) { return e1.age.compareto(e2.age); });
        }

    }

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